Basic Mathematics

& Logarithm
1
SECTION - A : BASIC MATHS Remarks
NUMBER SYSTEM 1. ‘1’ is neither prime nor composite.
2. ‘2’ is the only even prime number.
Natural Numbers
The counting numbers 1, 2, 3, 4 ........ are called Natural Composite Number
Numbers. The set of natural numbers is denoted by N. Let ‘a’ be a natural number, ‘a’ is said to be composite if
Thus, N = {1, 2, 3, 4,......}. N is also denoted by + or Z+ it
has atleast three distinct factors.
Whole Numbers Co-prime Numbers
Natural numbers including zero are called whole numbers. Two natural numbers (not necessarily prime) are coprime,
The set of whole numbers, is denoted by W. Thus W = if their H.C.F.(Highest common factor) is one.e.g. (1, 2),
{0, 1, 2,......}. W is also called as set of non-negative (1, 3), (3, 4), (3, 10), (3, 8), (5, 6), (7, 8) etc.
integers.
These numbers are also called as relatively prime
Integers numbers.
The numbers..– 3, – 2, –1, 0, 1, 2, ... are called
integers and the set is denoted by  or Z. Remarks
Thus (or Z) = {...–3, –2, –1, 0, 1, 2, 3.........} 1. Numbers which are not prime are composite
1. Set of positive integers, denoted by  + and numbers (except 1)
consists of {1, 2, 3, ...........} 2. ‘4’ is the smallest composite number.
3. Two distinct prime numbers are always co-prime but
2. Set of negative integers, denoted by – and
converse need not be true.
consists of {..........., –3, –2, –1}
4. Consecutive numbers are always co-prime
3. Set of non-negative integers {0, 1, 2, 3,...........}
numbers.
4. Set of non-positive integers {...., –3, –2, –1, 0}
Twin Prime Numbers
Even Integers
If the difference between two prime numbers is two, then
Integers which are divisible by 2 are called even integers. the numbers are called as twin prime numbers.
e.g. 0, ± 2, ± 4,..... eg. {3, 5}, {5, 7}, {11, 13}, {17, 19}, {29, 31}
Odd Integers
Rational Numbers
Integers which are not divisible by 2 are called as odd All the numbers those can be represented in the form p/q,
integers. e.g. ± 1, ± 3, ....... where p and q are integers and q  0, are called rational
Prime Number numbers and their set is denoted by Q.
Let ‘p’ be a natural number, ‘p’ is said to be prime if it has p
Thus, Q = { q : p, q  and q  0}. It may be noted that
exactly two distinct factors, namely 1 and itself.
e.g. 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31,...... every integer is a rational numbers.

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 Basic Mathematics & Logarithm

Irrational Numbers
2(3 + 5 + 2 2)( 5 – 1)
There are real numbers which cannot be expressed in p/ 
( 5 +1) × ( 5 – 1)
q form. These numbers are Called irrational numbers and
their set is denoted by Qc or Q’. 2(2 + 2 5 + 2 10 – 2 2)
=
(i.e. complementary set of Q) e.g. 2 , 1 + 3 , e,  etc. 4
Irrational numbers can not be expressed = 1+ 5 + 10 – 2
as recurring decimals.
Remark :
Example 3
1. e.  2.71 is called Napier’s constant and   3.14.
SURDS Find the square root of 7 + 2 10
If a is not a perfect nth power, then n a is called a surd of
the nth order. Sol. Let 7 + 2 10 = x + y .
a Squaring, x + y + 2 xy = 7 + 2 10
In an expression of the form , the denominator
b+ c Hence, x + y = 7 and xy = 10.These two
can be rationalized by multiplying numerator and the
relation give
denominator by b  c which is called the conjugate
x = 5, y = 2. Hence 7 + 2 10 = 5 + 2
of b  c . If x + y = a + b where x, y, a, b are
rationals, then x = a and y = b. Example 1 Remark :

Example 1 1. symbol stands for the positive square root

only.
Prove that log3 5 is irrational.
Example 4
Sol. Let log3 5 is rational.

p Prove that 3
2 cannot be represented in the
log3 5 =
q ; where p and q are co-prime
form p + q,
numbers.
where p and q are rational (q > 0 and is not
3p/q = 5  3p = 5q. which is not possible,
a perfect square).
hence our assumption is wrong and log3 5
is irrational. Sol. Put 3
2 =p+ q. Hence 2 = p3 + 3pq +

(3p2 + q) q , Since q is not a perfect square, it

Example 2 must be 3p2 + q = 0, which is impossible.
12 Real Numbers
Simplify :
3+ 5 – 2 2 The complete set of rational and irrational numbers is the
Sol. The expression =
set of real numbers and is denoted by R. Thus,
12(3 + 5 + 2 2) 12(3 + 5 + 2 2) R = Q  Qc. Real numbers can be represented as points
2 2
=
(3 + 5) – (2 2) 6+6 5 of a line. This line is called as real line or number line.

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 Basic Mathematics & Logarithm

All the real numbers follow the order property i.e. if there 5. A number is divisible by 6 iff the digit at the
are two distinct real numbers a and b then either unit place of the number is divisible by 2 & sum
a < b or a > b of all digits of the number is divisible by 3.
6. A number is divisible by 8 iff the last 3 digits,
all together, is divisible by 8.
7. A number is divisible by 9 iff sum of all it’s
digits is divisible by 9.
Remarks 8. A number is divisible by 10 iff it’s last digit is 0.
1. Integers are rational numbers, but converse need
not be true. 9. A number is divisible by 11 iff the difference
between the sum of the digits at even places and
2. Negative of an irrational number is an irrational sum of the digits at odd places is a multiple of 11.
number. Example. 1298, 1221, 123321, 12344321,
3. Sum of a rational number and an irrational number 1234554321, 123456654321, 795432
is always an irrational number e.g. 2 + 3 Example 5
4. The product of a non zero rational number & an
irr. number will always be an irrational number. Prove that :
1. The sum ab+ba is multiple of 11.
5. If a  Q and b  Q, then ab is rational number 2. A three-digit number written by one and the
only if a = 0. same digit is entirely divisible by 37.
6. Sum, difference, product and quotient of two Sol. 1. ab+ba = (10a + b) + (10b + a) = 11(a + b);
irrational numbers need not be an irrational number 2. aaa = 100a + 10a + a = 111a = 37.3a
(it may be a rational number also).
Example 6
Complex Number
If the number A 3 6 4 0 5 4 8 9 8 1 2 7 0 6 4 4
A number of the form a + ib is called complex number, B is divisible by 99 then the ordered pair
where a, b  R and i = 1 . Complex number is usually of digits (A, B) is
denoted by C. Sol. Sodd = A + 37 ; SEven = B + 34  A  B + 3= 0 or 11
and A + B + 71 is a multiple of 9
Remark
 A  B =  3 or 8 and A + B = 1 or 10
1. It may be noted that N  W  Q  R  C.
Ans. : (9, 1)
DIVISIBILITY TEST
1. A number is divisible by 2 iff the digit at the Example 7
unit place is divisible by 2. Consider a number N = 2 1 P 5 3 Q 4. Find
the number of ordered pairs (P, Q) so that the
2. A number is divisible by 3 iff the sum of the digits number ‘N is divisible by 44, is
of a number is divisible by 3.
Sol. Sodd = P + 9, SEven = Q + 6  Sodd – SEven
3. A number is divisible by 4 iff last two digits =P–Q+3
of the number together are divisible by 4. ‘N’ is divisible is 11 if P – Q +3 = 0, 11
P – Q = –3 ....(i)
4. A number is divisible by 5 iff digit at the unit or P – Q = 8 ....(ii)
place is either 0 or 5.

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 Basic Mathematics & Logarithm
3. a2 – b2 = (a + b) (a – b)
N is divisible by 4 if Q = 0, 2, 4, 6, 8 4. (a + b)3 = a3 + b3 + 3ab (a + b)
From Equation (i)
5. (a – b)3 = a3 – b3 – 3ab (a – b)
Q = 0 P = –3 (not possible)
Q = 2 P = –1 (not possible) 6. a3 + b3 = (a + b)3 – 3ab (a + b) = (a + b) (a2 + b2 – ab)
Q=4 P=1 Q=6 P=3 Q=8 P=5 7. a3 – b3=(a – b)3 + 3ab (a – b)=(a – b)(a2+ b2 + ab)
Number of ordered pairs is 3 8. (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
From equation (ii)
Q = 0 P = 8 Q = 2 P = 10 (not possible) 1 1 1
= a2 + b2 + c2 + 2abc  + + 
similarly Q  4, 6, 8 a b c
 No. of ordered pairs is 1
Total number of ordered pairs, so that number 1
9. a2 + b2 + c2 – ab – bc – ca = [(a – b)2 + (b –
‘N’ is divisible by 44, is 4 2
LCM AND HCF c)2 + (c – a)2]
1. HCF is highest common factor between any two 10. a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 –
or more numbers (or algebraic expression).
1
LCM is least common multiple between any two ab – bc – ca) = (a + b + c) [(a – b)2 + (b – c)2
2. 2
or more numbers (or algebraic expression)
+ (c – a)2]
3. Multiplication of LCM and HCF of two numbers If a + b + c = 0 then a3 + b3 + c3 = 3abc
is equal to multiplication of two numbers. 11. a4 – b4 = (a + b) (a – b) (a2 + b2)
a p   LCM of (a, p, ) 12. a4 + a2 + 1 = (a2 + 1)2 – a2 = (1 + a + a2) (1 – a + a2)
4. LCM of  b , q , m  = HCF of (b, q, m)
  Remarks
1 1 1
a p   HCF of (a, p, ) 1. ab + bc + ca = abc    
5. HCF of  b , q , m  = LCM of (b, q, m) a b c
  1
2. a2 + b2 + c2 – ab – bc – ca = [(a – b)2
2
6. LCM of rational and irrational number is not + (b – c)2 + (c – a)2]
defined.
Definition Of Indices
Remainder Theorem
If ‘a’ be any non-zero real or imaginary number and m is
Let P(x) be any polynomial of degree greater than or
positive integer than am = a.a.a. .........a (m times) where
equal to one and ‘a’ be any real number. If P(x) is divided
‘a’ is base ‘m’ is index.
by (x–a), then the remainder is equal to P(a).
Law of Indices
Factor Theorem
1. a0 = 1, (a  0)
Let P(x) be polynomial of degree greater than or equal to
1
1 and ‘a’ be a real number such that P(a) = 0, then (x – a) 2. a–m = m , (a  0)
a
is a factor of P(x). Conversely, if (x – a) is a factor of 3. am + n = am . an, where m and n real numbers
P(x), then P(a) = 0.
am
Some Important Identities 4. am – n = , where m and n real numbers, a  0
an
1. (a + b)2 = a2 + 2ab + b2 = (a – b)2 + 4ab
2. (a – b)2 = a2 – 2ab + b2 = (a + b)2 – 4ab 5. (am)n = amn 6. ap/q =
q
ap

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 Basic Mathematics & Logarithm

Example 8 In order to prove that f(x) is exactly divisible

by g(x), it is sufficient to prove that x – 1 and x
Find p and q so that (x + 2) and (x – 1) may be – 2 are factors of f(x). For this it is sufficient to
factors of the polynomial f(x) = x3 + 10x2 + px + prove that f(1) = 0 and f(2) = 0.
q.
Sol. Since (x + 2) is a factor  f(–2) must be zero Now, f(x) = 2x4 – 6x3 + 3x2 + 3x – 2
 f(1) = 2 × 14 – 6 × 13 + 3 × 12 + 3 × 1 – 2
 – 8 + 40 – 2p + q = 0 ...(1)
and, f(2) = 2 × 24 – 6 × 23 + 3 × 22 + 3 × 2 – 2
Since (x – 1) is a factor  f(1) must be zero
 f(1) = 2 – 6 + 3 + 3 – 2 and
 1 + 10 + p + q = 0 ...(2)
f(2) = 32 – 48 + 12 + 6 – 2
From (1) and (2), by solving we get p = 7 and
 f(1) = 8 – 8 and f(2) = 50 – 50
q = –18  f(1) = 0 and f(2) = 0
 (x – 1) and (x – 2) are factors of f(x)
Example 9  g(x) = (x – 1) (x – 2) is a factors of f(x).
Show that (2x + 1) is a factor of the expression Hence, f(x) is exactly divisible by g(x).
f(x) = 32x5 – 16x4 + 8x3 + 4x + 5.
 1 Example 12
Sol. Since (2x + 1) is to be a factor of f(x), f  – 
should be zero.  2
5 4 3 Using factor theorem, show that a – b, b – c
 1  1  1  1  1 and c – a are the factors of a(b 2 – c2 ) + b
f  –  = 32  –  –16  –  + 8  –  + 4  –  + 5
 2  2  2  2  2 (c2 – a2) + c (a2 – b2).
=0 Sol. By factor theorem, a – b will be a factor of the
Hence (2x + 1) is a factor of f(x).
given expression if it vanishes by substituting
Example 10 a = b in it. Substituting a = b in the given
expression,
Without using the Remainder theorem, find the we have a(b2 – c2) + b(c2 – a2) + c(a2 – b2)
remainder when f(x) = x6 – 19x5 + 69x4 – 151x3 + = b (b2 – c2) + b(c2 – b2) + c(b2 – b2)
229x2 + 166x + 26 is divided by x – 15. = b3 – bc2 + bc2 – b3 + c(b2 – b2) = 0
Sol. f(x) can be written as  (a – b) is a factor of a(b2 – c2) + b (c2 – a2)
(x6 – 15x5) – 4(x5 – 15x4 ) + 9(x4 – 15x3) – + c(a2 – b2).
16(x3 – 15x2) –11(x2 – 15x) + (x – 15) + 41 Similarly, we can show that (b – c) and (c – a)
or as f(x) = x5 (x – 15) – 4x4(x – 15) + 9x3(x – are also factors of the given expression.
15)–16x2(x – 15) – 11x(x – 15) + (x – 15) +41 Hence, (a – b), (b – c) and (c – a) are factors
Since the first six terms have x – 15 as a factor, of the given expression.
remainder = 41.
Example 13
Example 11
Show that x – 2y is a factor of 3x3 – 2x2 y –
Without actual division prove that 2x4 – 6x3 + 13xy2 + 10y3.
3x2 + 3x – 2 is exactly divisible by x2 – 3x + 2. Sol. Let f(x) = 3x3 – 2x2y – 13xy2 + 10y3
Sol. Let f(x) = 2x4 – 6x3 + 3x2 + 3x – 2 and g(x) = x2 Then f(2y) = 3(2y)3 – 2y(2y)2 – 13y2 (2y) +
– 3x + 2 be the given polynomials. 10y3 = 24y3 – 8y3 – 26y3 +10y3 = 0
Then g(x) = x2 – 3x + 2 = (x – 1) (x – 2) Hence x – 2y is a factor of f(x).

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 Basic Mathematics & Logarithm

Example 14
 x  ab   x  bc   x  ca 
Show that an – bn is divisible by a – b if n is any Sol.   c    a    b
positive integer odd or even.  ab   b c   c a 
Sol. Let a n – bn = f(a). By Remainder theorem, =0
f(b) = bn – bn = 0 (replacing a by b)  1 1 1 
 a – b is a factor of an – bn.  (x  (ab + bc + ca))    
a  b bc ca 
Example 15 =0
 x = ab + bc + ca.
Show that an – bn is divisible by (a + b) when n
is an even positive integer but not if n is odd. 1 1 1
If   = 0,the given equation
Sol. Let an – bn = f(a). a b bc ca
Now f(–b) = (–b)n – bn = bn – bn = 0 becomes an identity & is true for all x  R
if n is even and hence a + b is a factor of an – bn
If n is odd, f(–b) = –bn – bn = – 2bn  0. RATIO

Example 16 1. If A and B be two quantities of the same kind,

then their ratio is A : B; which may be denoted by
If a + b + c = 0, prove that
a4 + b4 + c4 = 2(b2 c2 + c2 a2 + a2 b2) =1/2(a2 + b2 + c2 )2 A
Sol. We know that, the fraction (This may be an integer or fraction)
B
(a+b+c)2 = a2+b2+c2+2ab+2bc+2ca
 a2+b2+c2 = –2(ab+bc+ca) [ a+b+c=0] 2. A ratio may represented in a number of ways e.g.
Squaring both sides of the relation
(a2 + b2 + c2)2 = [–2(bc + ca + ab)]2 a ma na
= 4{b2 c2 + c2 a2 + a2 b2} + 2 {bc. ca + ca. ab + ab. = = =....where m,n,......are non-zero
b mb nb
bc},
= 4(b2 c2 + c2 a2 + a2 b2) + 8abc (a + b + c) numbers.
= 4(b2 c2 + c2 a2 + a2 b2),since a + b + c = 0.
3. To compare two or more ratio, reduce them to
Therefore, 2(b2c2 + c2 a2 + a2b2)
=1/2 (a 2 + b2 + c2)2. common denominator.
Also (a2 + b2 + c2 )2 = (a4 + b4 +c4) + 2(b2c2 + 4. Ratio between two ratios may be represented as
c2 a2 + a2b2),
so that 4(b2c2 + c2 a2 + a2b2) = (a4 + b4 +c4) + a c a / b ad
2(b2c2 + c2 a2 + a2b2) :  =
b d c / d bc
 a4 + b4 +c4 = 2(b2c2 + c2 a2 + a2b2).
5. Ratios are compounded by multiplying them
Example 17
a c e ace
together i.e. . . ......= .......
x  ab x  bc x  ca b d f bdf
Solve the equation,  
ab bc 1 a 6. If a : b is any ratio then its duplicate ratio is a2 : b2;
1 1 1 triplicate ratio is a3 : b3..... etc.
=a + b + c. What happens if + +
a b b c ca
=0 7. If a : b is any ratio, then its sub-duplicate ratio is
a1/2 : b1/2, sub-triplicate ratio is a1/3 : b1/3 etc.

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 Basic Mathematics & Logarithm

PROPORTION Example 19
When two ratios are equal, then the four quantities
compositing them are said to be proportional. If a(y + z) = b(z + x) = c(x + y),
a c a–b b–c c–a
If = , then it is written as a : b = c : d or a : b : : c : d then show that x 2 – y 2 = y 2 – z 2 = z 2 – x 2
b d
1. ‘a’ and ‘d’ are known as extremes and 'b' and 'c' Sol. Given condition can be written as
are known as means.
y+z z+x x+y
2. An important property of proportion : Product of = = =k ....(1)
extremes = product of means. 1/ a 1/ b 1/ c
3. If a : b = c : d, then b : a = d : c (Invertando) (z + x) – (y + z) (x + y) – (x + z) (y + z) – (x + y)
= = =
1 1 1 1 1 1
4. If a : b = c : d, then a : c = b : d (Alternando) – – –
b a c b a c
a+b c+d
5. If a : b = c : d, then = (Componendo) x–y y–z z–x
b d = = = =k ....(2)
a–b b–c c–a
a–b c–d
6. If a : b = c : d, then = (Dividendo) ab bc ca
b d Multiplying (1) and (2), we get,
a+b c+d
7. If a : b = c : d, then= x 2 – y2 y2 – z2 z2 – x 2
a -b c-d = =
(Componendo and dividendo) a–b b–c c–a
a c e a + c + e + ..... a–b b–c c–a
8. If = = = ..... ,then each =  x 2 – y 2 = y 2 – z 2 = z 2 – x 2
b d f b + d + f + ......
Sum of the numerators
= Sum of the denominators Example 20

a c e xa + yc + ze + ..
9. If = = = .. ,then each = 2a + 3b + 2a – 3b
b d f xb + yd + zf + ..
If x = , show that 3bx2 –
1/ n 2a + 3b – 2a – 3b
a c e  xa n + yc n + zen 
10. If    .. ,then each =  n n n  4ax + 3b= 0.
b d f  xb + yd + zf 
x
Example 18 Sol. Taking the left hand side as , using
1
componendo and dividendo,
x+y y+z z+x
If = = , then find x : y : z.
2 3 4 x +1 2a + 3b
=
S um o f the nu m erators x –1 2a – 3b
Sol. Each = S um of the deno m in ato rs =
2(x + y + z) x + y + z (x +1)2 2a + 3b
= and therefore Squaring, = and again
9 9/2 (x – 1)2 2a – 3b
Each applying componendo and dividendo, We
= ( x  y 9z )  ( y  z )  (x  y 9z )  (x  z)  ( x  y 9z)  ( x  y)
3 4 2 x 2 +1 2a
2 2 2 get, = which gives the answer on
x y z 2x 3b
= = = x:y:z=3:1:5 cross multiplication.
3 / 2 1/ 2 5 / 2

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 Basic Mathematics & Logarithm

INTERVALS
Example 21 Intervals are subsets of R and generally its used to find
2y + 2z – x 2z + 2x – y 2x + 2y – z domain of inequality. If a and b are two real numbers
If = = , such that a < b then following types of intervals can be
a b c defined Open Interval (a, b) {x : a < x < b} i.e. extreme
then show that points are not included Closed Interval [a, b]{x : a  x 
9x 9y 9z b} i.e. extreme points are included It can possible when a
= = and b are finite Semi-Open Interval (a, b]{x : a < x  b}
2b + 2c – a 2c + 2a – b 2a + 2b – c
i.e. a is not included and b is included Semi-Closed Interval
9x 9y 9z [a, b) {x : a  x < b} i.e. a is included and b is not included
= =
2b + 2c – a 2c + 2a – b 2a + 2b – c
Method of Intervals
Sol. Since
2y + 2z – x 2z + 2x – y 2x + 2y – z Let F(x) =  x - a1 k  x  a 2 k .......  x  a n 1 k  x  a n k .
1 2 n 1 n

= = , Here k1, k2 ...., kn  Z are a1, a2,....an are fixed real num-
a b c
each ratio is equal to bers satisfying the condition
a1 < a2 < a3 < ... < an – 1 < an
For solving F(x) > 0 or F(x) < 0, consider the following
2(2z + 2x – y) + 2(2x + 2y – z) - (2y + 2z – x) algorithm:
2b + 2c – a 1. Mark the numbers a 1 , a 2 , .....a n on the number
9x axis and put plus sign in the interval on the right
= of the largest of these numbers,i.e. on the right of an.
2b + 2c – a
9y 2. Then put plus sign in the interval on the left of an
Similarly, Each ratio = and if kn is an even number and minus sign if kn is an
2c + 2a – b odd number. In the next interval, we put a sign
9z according to the following rule :
. Hence proved
2a + 2b – c 3. When passing through the point a n – 1 ,the polyno
mial F(x) changes sign if kn – 1 is an odd number.
Example 22 Then consider the next interval and put a sign
in it using the same rule.
2+x + 2– x 4. Thus, consider all the intervals. The solution of
, Solve : =2 the inequality F(x) > 0 is the union of all intervals
2+x – 2– x in which we put plus sign and the solution of the
2 inequality and F(x) < 0 is the union of all intervals in
Sol. Writing the R.H.S. as and using componendo which we put minus sign.
1
and dividendo, Frequently Used Inequalities
( 2 + x + 2 – x) + ( 2 + x – 2 – x ) 2 +1 1. (x –a) (x – b) < 0  x   a, b  . where a < b
=
( 2 + x + 2 – x) – ( 2 + x – 2 – x ) 2 –1 2. (x – a)(x – b) > 0  x    ,a    b,   , where a < b
3. x 2  a 2  x    a, a 
2+x 3
(i.e.) = 4. x 2  a 2  x  ( , a]  [a, )
2–x 1
2+x 9 5. ax2 + bx + c < 0, (a > 0)  x    ,   , where 
On squaring, We get, = and again (<) are the roots of the equation ax2 + bx + c = 0
2– x 1
applying componendo and dividendo, 6. ax2 + bx + c > 0, (a > 0)  x   ,     ,   ,
4 10 8 where ( < ) are the roots of the equation ax2 +
We get, = x = bx + c = 0
2x 8 5

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 Basic Mathematics & Logarithm

SECTION - B : LOG & PROPERTIES Example 23

LOGARITHM OF A NUMBER
The logarithm of the number N to the base ‘a’ is the 2-
log 5 13
 1  2log 5 9
exponent indicating the power to which the base ‘a’ must Compute  
 27 
be raised to obtain the number N.
This number is designated as logaN. log 513 1
Sol.  log 913
Hence, ogaN = x  ax = N, a > 0, a  1 & N >0 2log 5 9 2
Common and natural logarithm 2
log 513 log 513
 1  2 log 59 1  1  2 log 5 9

og10N is referred as a common logarithm and ogeN is     

27  27 
 27 
called as natural logarithm of N to the base Napierian
and is popularly written as n N. Note that e is an irrational 1
1
log 913 3
quantity lying between 2.7 to 2.8. Note that en x = x. 
27
  27  2
 3 3  3 8
log 313

The existence and uniqueness of the number ogaN follows 3 3 3

from the properties of an exponential functions.  3  3  13 8  3 2
 1 3 16
From the definition of the logarithm of the number N to
the base ‘a’, we have an identity : Example 23
a loga N  N, a  0, a  1 & N  0 a 3

This is known as the Compute logab if logab a = 4.

b
FUNDAMENTAL LOGARITHMIC IDENTITY. Sol. By the laws of logarithms, we have,
Other popular Identities 3
a 1 1 4 1
logab = log a – log b = - log b
oga1 = 0 (a > 0, a  1) b 3 ab
2 ab
3 2 ab

ogaa = 1 (a > 0, a  1) Also, 1 = logab ab = logab a + logab b = 4 +

og1/a a = –1 (a > 0, a  1) logab b
It follows that logab b = –3 and so
Remember
og10 2 = 0.3010,og103 = 0.4771, n 2 = 0.693, n 10 = 2.303
3
a 4 1 17
logab = – . (–3) =
The principal properties of logarithms : b 3 2 6
Let M & N are arbitrary positive numbers, a > 0, a  1,
Example 25
b > 0, b  1 and  is any real number then ;
1. oga(M.N) = ogaM + ogaN
1 1
2. oga(M/N) = ogaM – ogaN Compute the value of + .
3. ogaM = . ogaM log 2 36 log 3 36
1 1 1
4. og a M  oga M Sol. + = log362 + log363
 log 2 36 log 3 36
oga M 1
5. ogbM = (base change theorem) = log366 =
oga b 2
Remarks Example 26
1
1. ogba.ogab = 1  ogba =
 og a b Find the domain of logx–3(2x – 3).
2. ogba. ogcb. ogac = 1 Sol. x – 3 > 0, x – 3  1 and 2x – 3 > 0 x > 3,
3. ogyx. ogzy. ogaz = ogax. x  4 and x > 3/2 (3, 4)  (4, )
e ln a  a x .
x
4.
Motion Education | 394 - Rajeev Gandhi Nagar |): 1800-212-1799 | url : www.motion.ac.in | Page No. # 13
 Basic Mathematics & Logarithm

Example 27 Example 29

2 Compute log6 16 if log12 27 = a

Given log2a = s, log4b = s2 and logc2 (8) = .
3
s 1 4 4
Sol. log6 16 = 4 log6 2 = log 6 = 1 + log 3
2 2
a 2 b5
Write log 2 as a function of 's' (a, b, c > 0, c 1). Also, log12 27=a = 3 log12 3
c4
Sol. Given log2a = s ....(1) 3 3 3 3log 2 3
= = = =
log2b = 2s2 ....(2) log 3 12 1 + 2log 3 2 1 + 2 2 + log 2 3
log 2 3
s3  1
log8c2 = ....(3)
2
2a
 log23 =
2 log c s3  1 3–a
 3log 2 = (note that, obviously, a  3).
2
 4 log2c = 3(s3 + 1) ....(4) 4(3 – a)
 log6 16 = .
Hence the value of 2 log2a + 5 log2b – 4 log2c 3+ a
 2s + 10s2 – 3(s3 + 1)
SECTION - C

Example 28 LOGARITHMIC EQUATIONS

oga x = oga y possible when x = y i.e.
If log25 = a and log 225 = b, then find the value oga x = oga y  x = y
Always check the validity of the given equation i.e.
  1 2   1  x > 0, y > 0, a > 0, a  1
of log     + log   in terms of a
 9    2250  Example 30
 
and b. For x  0, what is the smallest possible value
Sol. log 25 = a; log 225 = b of the expression log(x3 – 4x2 + x + 26) –
2 log 5 = a ; log(25 · 9) = b log(x + 2) ?
or log 25 + 2 log 3 = b
 2 log 3 = b – a (x 3  4x 2  x  26)
Sol. log
2 (x  2)
1  1 
Now, log   + log 
9  2250 
(x 2  6x  13)(x  2)
= – 2 log 9 – log 2250 = log
(x  2)
= – 4 log 3 – [log 225 + log 10]
= – 2 (b – a) – [b + 1] = log (x2 – 6x + 13) [ x  – 2]
= – 2b + 2a – b – 1 = log{(x – 3) + 4}
2

= 2a – 3b – 1  Minimum value is log 4 when x = 3

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 Basic Mathematics & Logarithm

Example 31 Example 33
If log6 15 =  and log12 18 =  then compute the
value of log25 24 in terms of  & .   
If log 2 log 2 ( log 3 x) = log 2 log 3 ( log 2 y) 
1  log3 5 2  log 3 2 = 0 then find the value of (x + y).
Sol.  = ;=
1  log3 2 1  2log3 2  
Sol. log 2 log 2 ( log 3 x) = 0
Let log3 2 = x and log3 5 = y
log2(log3x) = 1
1 + y =  (1 + x) (1)
log3x = 2 x = 9
2 + x =  (2x + 1) (2) Also, log 2 log 3 ( log 2 y)  = 0
2–β log3(log2y) = 1 log2y = 3
x = (3)
2β –1 y = 8
Putting this value of x in (1) x + y = 17
 (1  )  (2   1)
y=  (4) SECTION - D : LOG INEQUALITIES
2 1
STANDARD LOG INEQUALITIES
3x  1
Now log25 24 = . Substitute the value of 1. For a > 1 the inequality 0 < x < y & loga x < loga y
2y
are equivalent.
5–β 2. For 0 < a < 1 the inequality 0 < x < y &
x and y to get log25 24 = loga x >loga y are equivalent.
2α + 2 α β – 4β + 2
3. If a > 1 then loga x < p  0 < x < ap
Example 32 4. If a > 1 then logax > p  x > ap
5. If 0 < a < 1 then loga x < p  x > ap
Suppose that a and b are positive real numbers such 6. If 0 < a < 1 then logax > p  0 < x < ap
7 2
that log27a + log9b = and log27b + log9a = .
2 3
Find the value of the ab.
7 1 1 7
Sol. log27a + log9b =  log3a + log3b = ;
2 3 2 2
2 1 1 2 Remarks
log27b + log9a =  log3b + log3a =
3 3 2 3
1. If the number & the base are on one side of the
Adding both the equations, we get,
unity, then the logarithm is positive; If the
1 1 7 2 25 number and the base are on different sides of unity,
log3(ab) + log3(ab) = + =
3 2 2 3 6 then the logarithm is negative.
5 25 2. The base of the logarithm ‘a’ must not equal to unity.
log3(ab) =
6 6
3. For a non negative number ‘a’ & n  2, n  N
 log3(ab) = 5
 ab = 35 = 243
n
a = a1/ n

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 Basic Mathematics & Logarithm

Example 34 The integral part is called the characteristic and the

decimal part is called the mantissa. It should be noted
If log0.3 (x  1) < log0.09 (x  1), then x lies in the that, if the characteristic of the logarithm of N is p then
interval number of significant digit in N = p + 1 if p is the non
Sol. First we note that for the functions involved negative characteristic of log N. Number of zeros after
in the given inequality to be defined (x  1) decimal before a significant figure start is –p – 1
must be greater than 0, that is, x > 1.
Example 36
Now log 0.3 ( x  1)  log 0.09 ( x  1)
Let x = (0.15)20. Find the characteristic and
 log 0.3 ( x  1)  log ( 0.3) 2 ( x  1)
mantissa in the logarithm of x, to the base 10.
2
 log 0.3 ( x  1)  log 0.3 ( x  1) Assume log102 = 0.301 and log103 = 0.477.
 15 
(x  1)2 > x  1 Sol. log x = log(0.15)20 = 20 log  
[Note that the inequality is reversed because  100 
= 20[log 15 – 2]
the base of the logarithms lies between 0 and 1]
= 20[log 3 + log 5 – 2]
(x  1)2  (x  1) > 0
(x  1) (x  2) > 0 …(i)  10 
Since x > 1, = 20[log 3 + 1 – log 2 – 2]  log5  log 
 2
the inequality (i) will hold if x > 2.
= 20[– 1 + log 3 – log 2]
Hence x lies in the interval (2, ).
= 20[– 1 + 0.477 – 0.301]
Example 35 = – 20 × 0.824 = – 16.48
Hence characteristic = –17 and mantissa = 0.52
x log5 x  5 then x may belongs to
Sol. We have, (log5 x)2 > 1 SECTION - F
[taking log5 both sides] MODULUS EQUATIONS /INEQUALITIES
 log5 x < –1 or log5 x > 1
ABSOLUTE VALUE FUNCTION /
1 MODULUS FUNCTION :
x< or x > 5
5 A function y = |x| is called the absolute value function or
 1 Modulus function. It is defined as :
But x > 0  x   0,   (5, )
 5

SECTION - E
 x if x  0
y = x=  
CHARACTERISTIC & MANTISSA   x if x  0
The common logarithm of a number consists of two parts,
Remarks
integral and fractional, of which the integral part may be
1. |x| < a  –a < x < a (a>0)
zero or an integer (+ve or –ve) and the fractional part a 2. |x| > a  x < –a or x > a (a>0)
decimal, less than one and always positive.

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 Basic Mathematics & Logarithm

Example 37 Example 38

Solution of the equation |x + 1| – |x – 1| = 3 If x satisfies | x  1 |  | x  2 |  | x  3 |  6,

Sol. Given, |x + 1| – |x – 1| = 3 then
Case-I Sol. Let y = |x–1| + |x–2| + |x–3|
Case-I x < 1
x < –1 y = 1–x + 2–x + 3–x = 6–3x
–x–1+x–1=3 Case-II 1  x < 2
 –2 = 3, which is not possible y= x–1 + 2–x + 3–x = 4–x
Case-III 2  x<3
Case-II y = x–1 + x–2 + 3–x = x
–1  x < 1 Case-IV x  3
x+1+x–1=3 y = x–1 + x–2 + x–3 = 3x–6

3
y

 x (rejected as x  [  1,1) )
2

Case-III 6
3

x1
2
x
0 1 2 3 4

x + 1 – x + 1 = 3  2 = 3,
which is not possible.
There is no value of x which satisfied From Graph, one can see that the solution set
for y  6 is x  (  ,0]  [4,  )
the Given equation.

Motion Education | 394 - Rajeev Gandhi Nagar |): 1800-212-1799 | url : www.motion.ac.in | Page No. # 17
 Basic Mathematics & Logarithm

Exercise 1 (Level-A) JEE Main Level

Basic Maths Log Properties

1. The value of 0. 2¯
¯¯¯
34 is - 8. If log (log 7 2
(logπ x)) vanishes, then x
equals.
(A) 232

990
(B) 232

9990
(C) 232

900
(D) 232

9909
(A) π 2 (B) 4 (C) 49 (D) 1
[ C. 75.76%, I.C. 6.82%, U.A. 17.42% ] [ C. 80.00%, I.C. 12.00%, U.A. 8.00% ]

2. If , then find the value 9.

a c e 1 1
= = +
b d f 1+log a+log c 1+log a+log b
b b c c

of in terms of a and b
4 2 2 2 4

is equal to-
2a b +3a c −5e f 1
6 2 2 5
+
2b +3b d −5f 1+log b+log c
a a

(C) 0 (D) 1
2 4 3 6

(A) a

b
2
(B) a

b
4
(C) a

b
3
(D) a

b
6
(A) abc (B) 1

abc

[ C. 67.01%, I.C. 28.99%, U.A. 4.00% ] [ C. 71.45%, I.C. 16.35%, U.A. 12.20% ]

3. If a, b, c are real and distinct numbers, 10. The value of

is
3 3 3
(1/log 3) log 36 4/log 9

then the value of

(a−b) +(b−c) +(c−a)
81 5
+ 27 9
+ 3 7

(a−b)(b−c)(c−a)
equal to
is
(A) 49 (B) 625
(A) 1 (B) abc (C) 2 (D) 3
(C) 216 (D) 890
[ C. 65.82%, I.C. 29.87%, U.A. 4.31% ]
[ C. 61.78%, I.C. 32.00%, U.A. 6.22% ]
3 2
4. If x – a is a factor of x – a x + x + 2, A B C
then ‘a’ is equal to 11. If 4 + 9 = 10 , where A = log164, B
= log3 9 & C = logx 83, then the value
(A) 0 (B) 2 (C) – 2 (D) 1 of x is.
[ C. 65.60%, I.C. 29.24%, U.A. 5.16% ]
(A) 11 (B) 10 (C) 9 (D) 12
[ C. 61.71%, I.C. 34.20%, U.A. 4.09% ]
5. Express 0 .3 6̄ as a fraction in
simplest form.
12. Simplify the expression
5 7 7 3
log log log log
(A) 11

30
(B) 11

90
(C) 33

10
(D) 10

33
7 3
+ 3 5
–5 3
–7 5

[ C. 65.26%, I.C. 17.82%, U.A. 16.92% ] (A) 1 (B) 0

(C) 7log35 (D) 5
6. If H.C.F. (a, b) = 12 and a × b = 1800,
then L.C.M. (a, b) = [ C. 58.94%, I.C. 34.64%, U.A. 6.42% ]

(A) 3600 (B) 900 (C) 150 (D) 90 13. The value of the expression
[ C. 63.12%, I.C. 33.42%, U.A. 3.46% ] +
2
is equal to
6
3
6
log (2000) log (2000)
4 5

7. If a, b, c are real, then a (a – b) + b (b (A) 1 (B) 1/2 (C) 1/3 (D) 1/6
– c) + c (c – a) = 0, only if [ C. 58.47%, I.C. 18.64%, U.A. 22.89% ]

(A) a + b + c = 0 14.
1
+
1
+
1
has
log abc log abc log abc

(B) a = b = c
√bc √ca √ab

the value equal to

(C) a = b or b = c or c = a
(A) 1/2 (B) 1 (C) 2 (D) 4
(D) a – b – c = 0 [ C. 58.46%, I.C. 35.51%, U.A. 6.03% ]
[ C. 62.50%, I.C. 31.70%, U.A. 5.80% ]

Motion Education | 394 - Rajeev Gandhi Nagar |): 1800-212-1799 | url : www.motion.ac.in | Page No. # 18
 Basic Mathematics & Logarithm
a b e c d
15. Let 3 = 4, 4 = 5, 5 = 6, 6 = 7, 7 = 22. x = 10x then the value of x is
1+log x

f
8, and 8 = 9. The value of the product
(abcdef), is
(A) 10 (B) 11 (C) 100 (D) 5
(A) 1 (B) 2 (C) √6 (D) 3 [ C. 66.39%, I.C. 11.04%, U.A. 22.57% ]
[ C. 56.26%, I.C. 36.66%, U.A. 7.08% ]

23. If log2(log3(log4(x))) = 0 and

16. If a = log12 18 & b = log24 54, then log3(log4(log2(y))) = 0 and
find the value of ab + 5(a – b). log4(log2(log3(z)))=0 then the sum of
(A) 0 (B) 1 (C) 2 (D) 3 x, y and z is
[ C. 55.20%, I.C. 40.36%, U.A. 4.44% ] (A) 89 (B) 58 (C) 105 (D) 50
[ C. 64.65%, I.C. 11.63%, U.A. 23.72% ]
17. is equal to
2
(log 4)
2 2

24. If log4 2+log4 4+log4 x+log4 16 = 6,

then x =
(A) 2 (B) 4 (C) 8 (D) 16
[ C. 53.48%, I.C. 41.78%, U.A. 4.74% ] (A) 64 (B) 4 (C) 8 (D) 32
[ C. 64.32%, I.C. 26.29%, U.A. 9.39% ]
18. If x = α satisfies the equation
x = 16, than α is
log√ (x−2)
x
25. If log4 log3 log2 x = 0. Then x =
(A) even composite
(B) odd composite (A) 16 (B) 8 (C) 4 (D) 12
[ C. 63.78%, I.C. 25.54%, U.A. 10.68% ]
(C) rational but not integer
(D) irrational 26. If log3 x + log3 y = 2 + log3 2 and
[ C. 53.20%, I.C. 39.20%, U.A. 7.60% ] log3 (x + y) = 2 then

19. Anti logarithm of 0.75 to the base 16 (A) x = 1, y = 8 (B) x = 8, y = 1

has the value equal to (C) x = 3, y = 6 (D) x = 9, y = 3
[ C. 60.57%, I.C. 12.00%, U.A. 27.43% ]
(A) 4 (B) 6 (C) 8 (D) 12
27. If log
[ C. 52.99%, I.C. 42.06%, U.A. 4.95% ] 2
√2
√x + log2 x + log4 (x ) +
3 4

Log Equations log8 (x ) + log16 (x ) = 40

then x is equal to
20. Solution set of the equation log(8 − (A) 8 (B) 16 (C) 32 (D) 256
10x − 12x2) = 3log(2x − 1) is
[ C. 60.12%, I.C. 36.90%, U.A. 2.98% ]

(A) {1} (B) {3, 2} (C) {5} (D) ϕ

[ C. 78.82%, I.C. 8.83%, U.A. 12.35% ] 28. If log log log
2 3 4
log5 (A) = x , then the
value of A is
21. Find x which satisfy the equation (A) 120x
logx–13 = 2
(B) 260x
(A) (1 + √3) (B) 2
x
5
4

(C) 2 3

(C) √3 (D) 1-√3

x
2
3

(D) 5 4

[ C. 68.08%, I.C. 15.96%, U.A. 15.96% ] [ C. 58.77%, I.C. 40.76%, U.A. 0.47% ]

Motion Education | 394 - Rajeev Gandhi Nagar |): 1800-212-1799 | url : www.motion.ac.in | Page No. # 19
 Basic Mathematics & Logarithm
29. − 2x + 65} = 2, Find x 35.
2
2
log x−3logx+3
log {x
(5−x) < 1
logx−1

(A) 4 (B) −3 (C) −5 (D) 10 (A) [0, 10] (B) (0, 10)
[ C. 55.40%, I.C. 12.91%, U.A. 31.69% ] (C) (0, 100) (D) [0, 100]
[ C. 60.33%, I.C. 33.97%, U.A. 5.71% ]
Log Inequalities
Characteristic & Mantissa
30. x
log
5
x
> 5 then x may belongs to
40
36. Let ‘m’ be the number of digits in 3
and ‘p’ be the number of zeroes in 3– 40
(A) (0, ) 1
(B) (0, ∞)
5
after decimal before starting a
(C) (-∞, 5) (D) (1, ∞) significant digit the (m + p) is
[ C. 74.33%, I.C. 12.91%, U.A. 12.76% ]
(log3 = 0.4771)
31. log5 (3x − 1) < 1 (A) 40 (B) 39 (C) 41 (D) 38
[ C. 55.03%, I.C. 39.56%, U.A. 5.41% ]

100
(A) (1, 2) (B) ( , 2) 37. Number of digits in N = 6 (where
1

(C) (2, 3) (D) (1, 3) log 2 = 0.3010, log 3 = 0.4771)

[ C. 72.10%, I.C. 12.05%, U.A. 15.85% ]
(A) 77 (B) 78 (C) 79 (D) 80
[ C. 54.19%, I.C. 25.12%, U.A. 20.69% ]
32.
2
log (x − 5x + 6)> −1
0.5

38. How many digits are contained in the

(A) (1, 3) number 275 ?
(B) (1, 4) (A) 21 (B) 22 (C) 23 (D) 24
(C) (2, 3) [ C. 53.85%, I.C. 38.91%, U.A. 7.24% ]

(D) (1, 2) ∪ (3, 4)

[ C. 64.27%, I.C. 29.36%, U.A. 6.37% ] 39. The number of zeros immediately
after the decimal in 3 −100

33.
2

(A) 50 (B) 47 (C) 48 (D) 49

35−x 1
log ( )≥ −
1/4 x 2

[ C. 47.98%, I.C. 33.44%, U.A. 18.58% ]

(A) (−√35, √35)

40. Consider the nubmer N = 2 .
100
log 2
4

(B) (−7, 0) ∪ (3, ∞) Then Number of digits of N before a

(C) [−7, −√35) ∪ [5, √35) decimal starts is [Given log102 =
0.3010]
(D) (−∞, −√35) ∪ (0, √35) (A) 14 (B) 15 (C) 16 (D) 17
[ C. 62.67%, I.C. 29.84%, U.A. 7.49% ] [ C. 47.83%, I.C. 18.72%, U.A. 33.45% ]

34.
2
log
3x+5
(9x + 8x + 8)> 2 Modulas Equation/ Inequalities

41. then x ∈ R
2
x −7x+12

2
> 0
2x +4x+5
(A) [− (B) (
4 −17 −4 −17
, ] , )
3 22 3 22

(C) [− 4
,
−17
) (D) (− 4
,
−17
] (A) (3,4) (B) (−∞, 3) ∪ (4, ∞)
3 22 3 22

[ C. 61.20%, I.C. 28.09%, U.A. 10.70% ] (C) (−∞, 4] (D) (3, ∞)

[ C. 76.88%, I.C. 14.21%, U.A. 8.91% ]

Motion Education | 394 - Rajeev Gandhi Nagar |): 1800-212-1799 | url : www.motion.ac.in | Page No. # 20
 Basic Mathematics & Logarithm
4 3 2
42. (x − 1)(x − 3) > 0
47. If f(x) = x − 2x + 3x − ax + b is a
polynomial such that when it is
divided by (x − 1) and (x + 1) the
(A) (−∞, 1)
remainders are 5 and 19 respectively.
(B) (3, ∞) If f(x) is divided by (x− 2), then
(C) (−∞, 1) ∪ (3, ∞) remainder is-
(D) [1, 3] (A) 0 (B) 5 (C) 10 (D) 2
[ C. 75.67%, I.C. 15.34%, U.A. 8.99% ]
[ C. 56.13%, I.C. 34.84%, U.A. 9.03% ]

43. Solve ∣∣x 48. If x, y, z are real numbers greater than

2
+ 4x + 3∣
∣ + 2x + 5 = 0

1 and 'w' is a positive real number. If

logxw = 24, logyw = 40 and logxyzw =
(C)
12 then logwz has the value equal to
(A) x = 4 (B) x = −4x = 1 + √3(D) x = 2
[ C. 71.89%, I.C. 16.59%, U.A. 11.52% ] (A) 1

120
(B) 2

120

2 2
(C) 3
(D) 5

44. | x | – | x | + 4 = 2x – 3 | x | + 1, then
120 120

[ C. 54.36%, I.C. 36.24%, U.A. 9.40% ]

find x?
(A) 3 (B) 2 (C) 0 (D) 1 49. The value of the expression
0 0
log (tan 6 ) + log (tan 12 ) +
10 10
[ C. 66.46%, I.C. 11.08%, U.A. 22.46% ] 0 0
log (tan 18 ) +. .... + log (tan 84 )
10 10

5 is
45. For the inequality (x + 5) (2x – 3) (–
x + 7)3(3x+8)2 < 0, x belongs to (A) a whole number
(B) an irrational number
(A) x ∈ (−5,
−8 8 3
) ∪ (− , ) ∪ (7, ∞)
(C) a negative integer
3 3 2

(B) x ∈ (−5,
−8 8 3

3
) ∪ (−
3
,
2
) ∪ (6, ∞)
(D) a rational number which is not an integer
[ C. 52.69%, I.C. 41.85%, U.A. 5.47% ]
(C) x ∈ (−5,
−8 8
) ∪ (− , 2) ∪ (7, ∞)
3 3

(D) x 50. If a, b, c ∈ N are three consecutive

8 3
∈ (−∞, − )∪( , ∞)
3 2

[ C. 60.11%, I.C. 28.19%, U.A. 11.70% ] terms of an increasing G.P. If

log a + log b + log c = 6 and
6 6 6

Mixed Problems (b − a) is a cube of a natural number

then (a + b + c) equals
46. Which of the following is correct ?
(A) 100 (B) 189 (C) 111 (D) 122
[ C. 51.61%, I.C. 16.94%, U.A. 31.45% ]
(A) x = y is called fundamental
log y
x

logarithmic identity.
(B) If x > 1 and 0 < y < 1 then log y
x > 0

(C) If 0 < x < 1 and y > 1 then log x

y > 0

(D) log 5 > log 5

6 4

[ C. 59.54%, I.C. 37.40%, U.A. 3.05% ]

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 Basic Mathematics & Logarithm

Exercise 1 (Level-B) Basic Learning

Basic Maths 10. Find the value of the expression

3 3
(log 2) + log 8. log 5 + (log 5)
1. Find the value of
49
(1−log
7
2)
+ 5
− log
5
4
.
11. (a) Which is smaller ? 2 or
(log 2 + log π)

2. Resolve the following into factors

π 2

(b) Prove that log3 5 and log2 7 are

(i) (x − y) − y
3

both irrational.
3

(ii) a − + 4
3

a
1
3

12. If log 3 = log

1
3 2
(iii) x − 6x + 11x − 6 log(4)+(1 + )
2x
3
(iv) x
2
− 9x − 10
2 2
x
(√3 + 27) . Find the value of x
(v) a (b − c)+b (c − a)+c (a − b)
(where x ∈ N )

3. Factorize :-
(i) 1 + x4 + x8 13. If loge log5 [√2x − 2 + 3]= 0 then
(ii) x4 + 4 find x.

4. Which is greater
(a) log 3 or log 5 14. (a) If x = log34 and y = log53, find the
2 1/2
value of log310 and log3(1.2) in terms
(b) log 11 or log
7 8
5 of x and y.
(b) If k = 16, find the value of
log
2
5

.
2

5. Simplify:(x
1/4 1/4 ( log 5)
+ y ) : k 2

2
3/2
√y
3
−1/2
x
((
y√ x
) + (
8
√y
3
) )
Log Equations

15. show that aa bb cc

loga logb logc
= =
b−c c−a a−b'

Log Properties =1
6. Calculate 4 5 log
4√2
(3−√6)−6 log (√3−√2)
8

16. If (log5 (x))

2
+ log 5x
(
5

x
) = 1, .then
find the value of x.
7. Simplify
(a) log 17. If x
4 3
√729√9−1 ⋅ 27−4/3
log x+5

1/3 3
= 10
5+log x
, then the value of
x is
(b) a

8. Given that log2 3 = a , log3 5= b , log7 2

18. (a) If log10 (x -12x+36)=2. Then the
2 = c , express the logarithm of the value of x is
number 63 to the base 140 in terms of
a , b & c. (b) If 9 1+log
− 3 − 210 = 0,
3
x 1+log
3
x

then the value of x is.

9. Prove that a where x =

x y

19. If log , then value

− b = 0 x
(3 − 8) = 2 − x
√loga b & y = √logb a , 3

of x =
(where a > 0, b > 0 & a, b ≠ 1).

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 Basic Mathematics & Logarithm
20. log4(log2x) + log2(log4x) = 2 then 28. Find the number of zeros after
find the value of x. decimal before a significant figure in
(i) 3-50 (ii) 2-100 (iii) 7-100
21. Simplify
Modulas Equation/ Inequalities

29. Solve the following linear equations

(i) | x | + 2 = 3
(ii) | x | – 2x + 5 = 0
22.
1
log ( )

Simplify 5 (iii) x| x | = 4
1
2 4
5
+log√ ( )
2
√7+√3

30. Solve the inequality:

+ log 1 ( )
2 10+2√21 |5 − 2x| < 1

23. If
31. Solve the inequalities
logb a. logc a + loga b. logc b + loga c. logb c = 3

(where a, b, c are different positive |2x − 4| < x − 1

real numbers ≠ 1), then find the value
of abc.
Mixed Problems
Log Inequalities 32. If
log (3)
4

24. Solve the inequality

log (3)
a
logb c = 3. 3 log4 3. 3 4

log (3)
log (3) 4
4
log (3)
log (5x − 1) > 0 .3 4
......∞
1/3

where a,b,c ∈ Q then the value of abc

is
25. Solve:
2x−6
log7 > 0
2x−1

33. If x = log 2a
(
bcd

2
), y = log3b (
acd

3
),

26. Solve the inequalities z = log4c (

abd

4
), w = log5d (
abc

5
)
2
log (x − 3x + 2)+1 < 0
1/6
Then Prove that
1 1 1 1
+ + + = log 120 + 1
x+1 y+1 z+1 w+1 abcd

Characteristic & Mantissa

27. Let x = (0. 15) . Find the 34. Prove that

20

characteristic and mantissa of the log5 √5√5√5. . . . . . . . . . . . . . . ∞ = 1

logarithm of x to the base 10.
Assume log 2 = 0. 301 and10

log 3 = 0. 477.
10

Motion Education | 394 - Rajeev Gandhi Nagar |): 1800-212-1799 | url : www.motion.ac.in | Page No. # 23
 Basic Mathematics & Logarithm

Exercise 2 JEE Advanced Level

Basic Maths (Single Correct) 7. If p, q ∈ N satisfy the equation

= (√x) then p & q are -
x
√x
x
1. The expression
x = log2 log9 (A) Prime Numbers (B) twin prime
√6 + √6 + √6+. . . . . . ∞
(C) Even number
(D) if log p is defined then log q is not &
simplifies to q p

vice versa
(A) 1 (B) (C) – 1 (D)
−1 1

2 2
[ C. 44.30%, I.C. 28.33%, U.A. 27.37% ]
[ C. 62.50%, I.C. 30.98%, U.A. 6.52% ]

8. The equation
2. If the polynomial (2x3 + ax2 + 3x – 5)
3 2
and (x + x – 2x + a) leave the same √1 + log √27 log x + 1 = 0
x 3
has
remainder when divided by (x – 2),
then the value of a is (A) one integral solution
(A) 2 (B) –2 (C) 3 (D) –3 (B) one irrational solution
[ C. 62.26%, I.C. 29.62%, U.A. 8.11% ] (C) two real solution
(D) no prime solution
3. If x = 2 + 2 + 2 then the value
2/3 1/3
[ C. 35.45%, I.C. 46.93%, U.A. 17.62% ]
of x − 6x + 6x is
3 2

(A) 3 (B) 2 (C) 1 (D) -2 9. Let M denote anti log 0.6 and N 32

denote the value of

[ C. 61.59%, I.C. 34.29%, U.A. 4.13% ]
49
(1−log
+ 5
7
2)
. Then M.N is :− log
5
4

(Multiple Correct) (A) 100 (B) 400 (C) 50 (D) 200

[ C. 34.97%, I.C. 61.63%, U.A. 3.40% ]
4.
1
k k k

If , then is
(a +b +c ) k
a b c
= =
(Multiple Correct)
d e f 1
k k k
(d +e +f ) k

equal to :(k∈n)
10. Let N = . Then N is-
log 135 log 5
3 3

3
−
(A) a

d
(B) b

e
(C) c

f
(D) a

d
2
log
15
3 log
405
3

[ C. 75.97%, I.C. 20.54%, U.A. 3.49% ]

(A) a natural number
Log Properties (Single Correct) (B) a prime number
(C) a rational number
5. Greatest integer less than or equal to
the number (D) an integer number
log (15) . log (2) . log ( ) is
1

2
1
3 6
[ C. 81.40%, I.C. 12.89%, U.A. 5.71% ]
6

(A) 4 (B) 3 (C) 2 (D) 1

11. Which of the following when
[ C. 49.26%, I.C. 46.44%, U.A. 4.30% ]
simplified, vanishes ?

6. Which one of the following denotes (A) 1

+
2
−
3

the greatest positive proper fraction ? log 2

3
log 4
9
log
27
8

(B) log ( ) + log ( )

2
2

3 4
9

(A) ( 1
)
log 6
2
(B) ( 1
)
log 5
3
(C) − log8 log4 log2 16
4 3

(
1
) (D) log10 cot 1° + log10cot 2° + log10 cot 3° +
(C) 3 −log 2
(D) 8 − log 2

......... + log10 cot 89°

3 3

[ C. 48.03%, I.C. 49.16%, U.A. 2.81% ]

[ C. 71.43%, U.A. 28.57% ]

Motion Education | 394 - Rajeev Gandhi Nagar |): 1800-212-1799 | url : www.motion.ac.in | Page No. # 24
 Basic Mathematics & Logarithm
12. Let a = log 3, b =
log 3
(all 16. The value of | A – B + C | is equal to
log(log 3)

logarithms on base 10) the number ab

is (A) – 30 (B) 32
(A) an odd integer (C) 28 (D) 30
(B) an even integer [ C. 0.77%, I.C. 28.84%, U.A. 70.39% ]

(C) a prime number

Log Equations (Single Correct)
(D) a composite number
[ C. 66.67%, I.C. 29.67%, U.A. 3.66% ] 17. Number of real solution of the
equation √log (−x) = log
10 10
√x2

13. Which of the following real numbers

are non-negative? (A) none (B) exactly 1
(C) exactly 2 (D) 4
4 −5 −3 [ C. 52.70%, I.C. 42.70%, U.A. 4.60% ]
(A) log 5
√
25.
√
8 3
.4 2

(B) log 7π (sin

5π
) 18. If x1 and x2 are solutions of the
equation
cos 6
4

(C) log tan

4π (cot
7π

6
)
log5 (log64 |x| + (25)
x
−
1
) = 2x ,
3 2

3
−5 −7
then
(D) log √ √
(A) x1 = 2x2 (B) x1 + x2 = 0
9. 27 3
. 243 5
2

[ C. 39.64%, U.A. 60.36% ]

(C) x1 = 3x2 (D) x1x2 = 64
[ C. 47.57%, I.C. 49.51%, U.A. 2.91% ]
Question No. 14 - 16

Paragraph: 19. Number of real solution (x) of the

equation |x − 3|
2

A denotes the product xyz where x, y

3x −10x+3
= 1 is

and z satisfy (A) exactly four (B) exactly three

log3 x = log5 – log7
(C) exactly two (D) exactly one
log5 y = log7 – log3
[ C. 44.15%, I.C. 50.85%, U.A. 5.00% ]
log7z = log3 – log5
B denotes the sum of square of 20. 2 log4 (4 − x) = 4 − log2 (−2 − x).
solution of the equation Find the value of x.
log2 (log2 x6 – 3) – log2 (log2 x4 – 5)
= log2 3 (A) −4 (B) 4 (C) 3 (D) -3
C denotes characteristic of logarithm [ C. 38.60%, I.C. 23.98%, U.A. 37.43% ]

log2 (log2 3) – log2 (log4 3) +

log2 (log4 5) – log2 (log6 5) + (Multiple Correct)
log2 (log6 7) – log2 (log8 7)
21. The equation logx2 16 + log2x 64 = 3
14. The value of log2A + log2B + log2C is has
equal to
(A) one irrational solution
[ C. 2.63%, I.C. 23.03%, U.A. 74.34% ]
(B) no prime solution
15. The value of A + B + C is equal to (C) two real solutions
(D) one integral solution
[ C. 79.25%, I.C. 18.04%, U.A. 2.71% ]
[ C. 1.79%, I.C. 26.86%, U.A. 71.35% ]

Motion Education | 394 - Rajeev Gandhi Nagar |): 1800-212-1799 | url : www.motion.ac.in | Page No. # 25
 Basic Mathematics & Logarithm
22. The equation Log Inequalities (Single Correct)
2 9

has
[ ( log x) − log x+5 ]

27. If log10 2 = 0.3010 and log10 3 =

3 2 3
x = 3√3

(A) exactly three real solution 0.4771, then the number of digits in
(B) at least one real solution 615 are-
(C) exactly one irrational solution (A) 15 (B) 12
(D) complex roots
[ C. 76.35%, I.C. 20.36%, U.A. 3.29% ]
(C) 13 (D) 11
[ C. 56.23%, I.C. 36.84%, U.A. 6.93% ]

23. For the equation log |x + y| =

1

2
28. Solve log(x + 3)(x – x) < 1
100 2

and log y − log |x| = log 4 ,

10 10 100

(x, y) is

(A) (10/3, 20/3) (B) (10, 20) (A) x ∈ [−3, −2] ∪ (−1, 0) ∪ (1, 2)
(C) (–10, 20) (D) (-10/3, –20/3) (B) x ∈ ( − 3 , − 2 ) ∪ ( − 1 , 0 )∪( 1 , 3)
[ C. 61.65%, I.C. 36.02%, U.A. 2.33% ] (C) x ∈ ( − 3 ,− 2 ) ∪ [ − 1 ,0 ) ∪ ( 1 , 2]
(D) x ∈ ( − 3 , − 2 ] ∪ ( − 1 ,0 ] ∪ ( 1 ,3]
24. If log7x + log13x = 1 and x = 13
log 7
k
[ C. 50.85%, I.C. 42.54%, U.A. 6.61% ]
then k is divisible by
(Multiple Correct)
(A) 7 (B) 13 (C) 17 (D) 119
29. If then
1
[ C. 56.02%, I.C. 9.52%, U.A. 34.46% ] ≤ log x ≤ 2
2 0.1

25.
8
log ( )

The equation = 3 has -

8 2

(A) The maximum value of x is 1/√10

x

2
(log x)
8

(B) x lies between 1/100 and 1/√10

(A) no integral solution
(B) one natural solution (C) x does not lie between 1/100 and 1/√10
(C) two real solution (D) The minimum value of x is 1/100
(D) one irrational solution [ C. 54.90%, I.C. 7.73%, U.A. 37.37% ]
[ C. 42.61%, I.C. 19.89%, U.A. 37.50% ]
30. Solution set of the inequality
(Column Match)
4
(log2 x) −
2

26. Match the following :

3
x 32
[log 1 ( )] + 9 log ( )<
8 2 x
2
2

4(log 1 x)
2

is (a, b) ∪ (c, d) then the correct

statement is
(A) a = 2b and d = 2c
(B) b = 2a and d = 2c
(C) logcd = logba
(D) there are 4 integers in (c, d)
[ C. 46.09%, U.A. 53.91% ]

(A) A → RS, B → PQS, C → P, D → Q

(B) A → R, B → PQR, C → Q, D → P
(C) A → P, B → PRS, C → S, D → R
(D) A → Q, B → QRS, C → R, D → S
[ C. 56.77%, I.C. 20.65%, U.A. 22.58% ]

Motion Education | 394 - Rajeev Gandhi Nagar |): 1800-212-1799 | url : www.motion.ac.in | Page No. # 26
 Basic Mathematics & Logarithm
Characteristic & Mantissa (Single 37. If x satisfies |x − 1| +|x − 2| +
Correct) |x − 3| ≥ 6, then

31. If p is the number of natural numbers (A) 0 ≤ x ≤ 4 (B) x ≤ -2 or x ≥ 4

whose logarithms to the base 10 have
the characteristic p and q is the (C) x ≤ 0 or x ≥ 4 (D) 1 ≤ x ≤ 3
number of natural numbers logarithms [ C. 46.05%, I.C. 48.27%, U.A. 5.68% ]
of whose reciprocals to the base 10
have the characteristic - q then
log10 p - log10 q has the value equal to 38. Number of solutions of |x − 1| + |x −
(A) p + q − 1 (B) p − q + 1 2| + |x − 4| + | x − 6| + |x − 7| = 2x +
(C) p + q + 1 (D) p + q 10 is
[ C. 45.43%, I.C. 48.47%, U.A. 6.10% ] (A) 0 (B) 1 (C) 2 (D) 3
[ C. 18.90%, I.C. 48.03%, U.A. 33.07% ]
32. Number of zeros after decimal before
a significant figure comes in the Mixed Problems (Single Correct)
number ( ) , is
100
5

39. The number of ordered pairs (x,

6

(A) 4 (B) 5 (C) 6 (D) 7 y) satisfying the equation x − y =

[ C. 21.74%, I.C. 48.55%, U.A. 29.71% ] (log2y – log2x) (2 + xy) and x3 + y3 =
16
(Multiple Correct)
(A) 1 (B) 2 (C) 3 (D) 0
33. Let N =
3 6
3 log 2−2 log(log 10 )+log((log 10 ) )
2
[ C. 33.33%, I.C. 62.55%, U.A. 4.12% ]

10
where base of the logarithm is 10. The 40. The number of integral solutions (x,
characteristic of the logarithm of N to y) of two equations logxy + logyx = y
the base 3, is not equal to and x2 + y = 6 is
(A) 2 (B) 3 (C) 4 (D) 5 (A) 0 (B) 1
[ C. 56.86%, I.C. 41.18%, U.A. 1.96% ] (C) 2 (D) Infinite
[ C. 15.99%, I.C. 23.39%, U.A. 60.62% ]
34. The number of odd positive integers,
the logarithm of whose reciprocal to
the base 3 have characterstic – 5, is (Multiple Correct)
less then
(A) 81 (B) 82 (C) 161 (D) 162 41. Which of the following statement(s)
[ C. 55.32%, I.C. 42.55%, U.A. 2.13% ]
is/are true ?
(A) log 2 lies between
1
and
1

Modulas Equation/ Inequalities (Single 10 4 3

Correct) (B) log cosec(

π
)
(cos
π

3
) = −1
6

35. |x – 3| + 2|x + 1| = 4 (C) e ln(ln 3)

is smaller than 1

(D) log 10
1 +
1

2
log10 3 + log10 (2 + √3)

(A) -1 (B) 1 (C) 0 (D) 5 = log10 (1 + √3 + (2 + √3))

[ C. 53.78%, I.C. 40.47%, U.A. 5.75% ]
[ C. 32.68%, I.C. 58.69%, U.A. 8.63% ]

36. If |x − 4| + |x| < 3 then

42. If x >1 >y >0 then the value of the
expression log ( )+ log ( ) can
x y

x y y x
(A) x ∈ [ 7
, 4) (B) x ∈ [ 7
, 4]
2 2
never be
(C) x ∈ ( 7

2
, 4) (D) x ∈ ϕ (A) 4 (B) 2 (C) 1 (D) 0
[ C. 47.48%, I.C. 20.45%, U.A. 32.07% ] [ C. 28.39%, I.C. 60.28%, U.A. 11.33% ]

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 Basic Mathematics & Logarithm

Exercise 3 Numerical Type

1. Let A denotes the value of 9. Find the square of the sum of the
⎛ ab+√(ab) −4(a+b)
2
⎞
roots of the equation.
log10 ⎜ ⎟+ log3x . log4 x . log5x= log3 x. log4
x+log4x . log5x + log5x . log3x
2
⎝ ⎠

2
⎛ ⎞
10. Let a and b be real numbers greater
ab−√(ab) −4(a+b)

log ⎜ ⎟
10

⎝
2
⎠
than 1 for which there exists a
positive real number c, different from
when a = 43 and b = 57 and B denotes
the value of the expression 1, such that
). Find the value
log 18 log 3 2(log c + log c) = 9 log c.
a b ab
(2 ) . (3
6 6

Find the largest possible value of

of (A · B). loga b.

2. If x = log2a a, y= log3a 2a and z =

11. If x , y > 0, logy x + logx y= and
10

log4a 3a, Find the value of

1+xyz
3

xy = 144, then = √N where N is

2 yz x+y

a natural number, find the value of N.

3. The value of log34.log45.log56.log
67.log78.log89 is equal to -
12. Let (x − 5) = 1 and sum of
(x−7)

possible values of x be λ then is λ

4. If 5
logx (logx)−1 (logx)+1 (logx)−1 13
−3 =3 −5

, (where Base of log is 10) then x

is
13. Find the value of x of the equation,
20

5
log
= 50 − x
10
x
is _______ . log
10
5

5. If log x
log18 (√2 + √8)=
1

3
. Then (Divide your answer by 10)
the value of 8x is equal to ______ .
14. is equal to
6. If a, b, c are positive real numbers
such that
= 49 and c
15. The value of x satisfying the equation
log 7 log 11 log 25
a = 27; b
3 7 11

= √11.
x
log (2 + x − 41) = x (1 − log 5)
10 10

The value of is _______ .

2 2 2
( log 7) ( log 11 ) ( log 25 )
(a 3
+ b 7
+ c 11
)
x
– 460 equals. 16. The solution of log2(4 × 3 − 6) −
log2(9x − 6) = 1 is
7. Number of real values of x satisfying
the equation
log (√x + 1) = log (1 − √x)is
17. The value of x satisfying the
2 3
equation
equal to
9
log (log
3 2
x)
= log2 x − (log2 x)
2
+ 1 is
____ .
8. Let y =
√log 3 ⋅ log2 12 ⋅ log2 48 ⋅ log2 192 + 16−
2

log2 12 ⋅ log2 48 + 10 18. Solve for x : log5120 + (x – 3) – 2.

Find y∈N log5(1–5x–3) = – log5(0.2 – 5x–4).

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 Basic Mathematics & Logarithm
19. If the value of x = k, then find the 24. If log3x45 = log4x40√3 then the
value of ∣∣ ∣∣ satisfying equation:
1

k
characteristic of x3 to the base 7 is
(6x + 23x + 21) = 4 –
2
log
(2x+3)

25. Let ABC be a triangle right angled at

2
log (3x+7)
(4x + 12x + 9)

C. The value of
log a+log a
x
20. If 5x ⋅ √8x−1 = 500, then the
b+c c−b
(b + c ≠ 1, c − b ≠ 1)
log a.log a
b+c c−b

value of x is. equals

21. If 3 − 2 x − 3 = 0, then the 26. If log2 (a + b) + log2 (c + d) ≥ 4. Then

2 log x
3

number of values of `x' satisfying the the minimum value of the expression
equation is a + b + c + d is

22. If x, y ∈ R satisfy simultaneously the 27. If x1 and x2 are the two solutions of
8
the equation 3 log2 x log 9

equation logx + =2
log(xy ) − 12x 16

2 2 3

, then find the value of

(logx) +(logy) 3
1
= log ( )
and 3 3

).
1 2 2
x
8 (x + x
log( ) 16 1 2

= 0. Compute
y xy
logy + 2 2
2
(logx) +(logy)

. 28. Find the integral value of x satisfying

the equation ∣∣log x − 2∣∣ − | log3x − √3

23. Given that log(2) = 0.3010 , then the 2 | = 2.

number of digits in the number
20002000 is
(Divide your answer by 100)

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 Basic Mathematics & Logarithm

Exercise 4 (Level-A) JEE Main (Previous Year Questions)

1. The number of real roots of the 6. For a natural number n, let

equation 5 +|2 − 1| = 2x (2x - 2) is :
x
α = 19 − 12 . then the value of
n
n n

is ........
31α9 −α10

(A) 2 (B) 1 (C) 3 (D) 4 57α8

[ C. 31.01%, I.C. 43.60%, U.A. 25.39% ] (JEE Main 2019) [ C. 11.98%, I.C. 17.05%, U.A. 70.97% ] (JEE Main 2022)

2. The sum of the solutions of the 7. 25

190
− 19
190
− 8
190
+ 2
190
is
equation divisible by
∣√x − 2∣+√x(√x − 4)+2, (x > 0)
(A) 34 but not by 14
is equal to :
(B) 14 but not by 34
(A) 12 (B) 9 (C) 10 (D) 4
(C) both 14 and 34
[ C. 24.01%, I.C. 54.61%, U.A. 21.38% ] (JEE Main 2019)
(D) neither 14 nor 34
3. The number of solutions of the [ C. 10.23%, I.C. 55.85%, U.A. 33.92% ] (JEE Main 2023)

equation log (x − 1) = log (x − 3)

8. If a + b + c = 1, ab + bc + ca = 2 and
4 2

is ______.
abc = 3, then the value of a4 + b4 + c4
[ C. 21.03%, I.C. 36.69%, U.A. 42.28% ] (JEE Main 2021)
is equal to __________.
4. The number of significant figures in [ C. 8.35%, I.C. 39.50%, U.A. 52.14% ] (JEE Main 2021)

50000.020 × 10–3 is _______.

9. The number of solutions of the
[ C. 14.21%, I.C. 43.16%, U.A. 42.63% ] (JEE Main 2021) equation log(x + 1) (2x2+7x+5) +
log(2x+ 5)(x+1)2–4 = 0, x > 0, is
5. The number of integral solutions x of ______.
2

is
x−7
log 7
( ) ≥ 0 [ C. 6.92%, I.C. 18.58%, U.A. 74.51% ] (JEE Main 2021)
(x+ ) 2x−3
2

(A) 5 (B) 7 (C) 8 (D) 6 10. Let a,b,c be three distinct positive real
[ C. 12.53%, I.C. 58.54%, U.A. 28.93% ] (JEE Main 2023) numbers such that
and
log a log b
e e
(2a) = (bc)
log 2 log c
b e
= a e

then 6a + 5bc is equal to

[ C. 1.24%, I.C. 30.71%, U.A. 68.05% ] (JEE Main 2023)

Motion Education | 394 - Rajeev Gandhi Nagar |): 1800-212-1799 | url : www.motion.ac.in | Page No. # 30
 Basic Mathematics & Logarithm

Exercise 4 (Level-B) JEE Advanced (Previous Year Questions)

1. The value of 2. Let (x0, y0) be the solution of the

following equations (2x)
ln ( 2 )
=
6 + log 3
/2

, Then x0
ln ( 3 ) ln ( x ) ln ( y )
(3y) , 3 = 2

is
(A) 1

6
(B) 1

3
(C) 1

2
(D) 6
is [ C. 28.57%, I.C. 37.23%, U.A. 34.20% ] (JEE Adv. 2013) [ C. 22.21%, I.C. 36.58%, U.A. 41.21% ] (JEE Adv. 2011)

3. The value of 1
1
log 7
4
2 log
2
( log
2
9)

((log 9) ) × (√7)
2

is_____
[ C. 10.39%, I.C. 18.10%, U.A. 71.51% ] (JEE Adv. 2018)

Motion Education | 394 - Rajeev Gandhi Nagar |): 1800-212-1799 | url : www.motion.ac.in | Page No. # 31
 Basic Mathematics & Logarithm
Answer Key
Exercise 1 (Level-A) JEE Main Level

1. A 2. B 3. D 4. C 5. A 6. C 7. B 8. A

9. D 10. D 11. B 12. B 13. D 14. B 15. B 16. B

17. D 18. A 19. C 20. D 21. A 22. A 23. A 24. D

25. B 26. C 27. D 28. D 29. C 30. A 31. B 32. D

33. C 34. B 35. B 36. B 37. B 38. C 39. B 40. C

41. B 42. C 43. B 44. A 45. A 46. A 47. C 48. B

49. A 50. B

Exercise 1 (Level-B) Basic Learning

1. 25

2
2. (i) (x − 2y) (x
2
− xy + y )
2

1 2 1 1
(ii) (a − + 1) (a + − a + + 2)
a a
2 a

(iii) (x − 1)(x − 2)(x − 3)

2
(iv) (x + 2) (x − 2x − 5)

(v) − (a − b)(b − c)(c − a)

3. 4.
(b) log
4 2 2 2
(i) (x − x + 1)(x − x + 1)(x + x + 1)
(a) log 3 > log 5 11 > log 5
2 1/2 7 8

(ii) (x 2
− 2x + 2)(x
2
+ 2x + 2)

5. xy
6. 9 7. (a) = -1 (b) log b
N

8. 1+2 ac

1+abc+2c
10. 1
11. (a) (log 2
π + logπ 2) > 2

12. ϕ 13. 3
14. xy+2 2y+xy−2 16. x = 1, 5,
1

(a) log (10) = , log (1. 2) = 25

3 2y 3 2y

17. x = 10
3
or x = 10
−5

(b) 625

18. (a) x = - 4, 16 (b) 5

19. 2 20. 16
21. 1 22. 6
23. 1 24. (
1

5
,
2

5
)

Motion Education | 394 - Rajeev Gandhi Nagar |): 1800-212-1799 | url : www.motion.ac.in | Page No. # 32
 Basic Mathematics & Logarithm
25. (−∞,
1
) 26. (−∞, −1)∪(4, ∞)
2

27. characteristic = -17 and mantissa = 0.52

28. (i) 23 (ii) 30 (iii) 84 29. (i) x = ±1 (ii) x = 5 (iii) x = 2
30. (2, 3) 31. (
5
, 3)
3

32. 16
Exercise 2 JEE Advanced Level

1. C 2. D 3. B 4. A,B,C 5. C 6. C 7. D 8. D

9. A 10. A,B,C,D 11. A,B,C,D 12. A,C 13. A,B,C 14. 5 15. 34 16. D

17. C 18. B 19. B 20. A 21. A,B,C,D 22. A,B,C,D 23. A,C 24. A,B

25. B,C 26. A 27. B 28. B 29. A,B,D 30. B,C 31. B 32. D

33. A,C,D 34. B,C,D 35. A 36. D 37. C 38. C 39. A 40. B

41. A,B,D 42. B,C,D

Exercise 3 Numerical Type

1. 12 2. 1 3. 2 4. 5 5. 1 6. 9 7. 1 8. 6

9. 3721 10. 2 11. 507 12. 1 13. 10 14. 1 15. 41 16. 1

17. 2 18. 1 19. 4 20. 3 21. 3 22. 5 23. 66.03 24. 2

25. 2 26. 8 27. 17 28. 9

Exercise 4 (Level-A) JEE Main (Previous Year Questions)

1. B 2. C 3. 1 4. 7 5. D 6. 4 7. A 8. 13

9. 1 10. 8
Exercise 4 (Level-B) JEE Advanced (Previous Year Questions)

1. 4 2. C 3. 8

Motion Education | 394 - Rajeev Gandhi Nagar |): 1800-212-1799 | url : www.motion.ac.in | Page No. # 33
POINTS TO REMEMBER
POINTS TO REMEMBER
POINTS TO REMEMBER