Comprehensive Oxford Mathematics and Physics Online School (COMPOS)

Year 12

Mathematics Assignment 08
Equations
Elena Boguslavskaya, Vladimir Chernov, Alexander Lvovsky

Due 12 May, 2024

This is the eighth Mathematics assignment from COMPOS Year 12. This assignment is designed to
stretch you and no student is expected to complete all questions on the first attempt. The problems
are hard, but do not let this discourage you. Give each problem a go, and skip to the next one if you
are stuck. The questions in each section are arranged in the order of increasing complexity, so you
should try all sections. Very similar problems will be discussed in tutorials and webinars, so think of
the questions you would like to ask. We hope that eventually you will be able to solve most of the
problems. Good luck!

Total 61 marks.

1 Irrational equations

1.1 Valid Values

When solving equations, it is important to establish the range of valid values of the unknown variable(s).

Example 1. Solve the equation: √ √

x−5= x−5
Solution: Squaring both sides, we get x − 5 = x − 5, equivalent to 0 = 0, which is true for any value of x.
However, the expression under a square root cannot be negative, so we must have x − 5 ≥ 0 ⇒ x ≥ 5

Answer: x ≥ 5, x ∈ R

As we see, solving an irrational equation may involve squaring both sides, which often leads to extraneous
solutions, which have to be checked against the range of valid values. Usually the best way of checking is
to substitute the roots into the original equation. The above example is an exception: the equation has
infinitely many roots, so it is impossible to check them by substitution.

When solving an equation, always perform a “sanity check” by substituting your answers into the
original equation!

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In this homework, we ask you to do the sanity check explicitly as a part of the solution you submit.

Example 2. Solve the equation: p

4 − x2 = −1
Solution: The left-hand side of the equation is positive for any value of x. Hence, there are no solutions.

Answer: x ∈ ∅.

Problem 1 (3 marks). Solve the equations:

√ √
a) x − 7 + 4 1 − x = 3;
√
b) (2x − x2 ) 5x − 3 = 0

Problem 2 (6 marks). Solve the equations:

√
a) 2x2 − 7x + 5 = 1 − x;
√
b) 10 − x − x2 − x − 1 = 0.

1.2 Monotonic Functions

You can use the properties of monotonic functions to solve equations. Consider the following example:

Example 3. Solve the equation: √ √

5x − 1 + 3x + 10 = 7

Solution.

This equation can be solved by squaring both sides, but you can also guess the solution by noticing that the
function in the left-hand side is monotonically increasing1 . The right-hand side is a constant y = 7, so the
graphs of the two sides can only have one point of intersection. That means that the equation can only have
one solution. We will use trial and error to find x = 2.
1A function f (x) is monotonically increasing if for any x1 , x2 such that x2 > x1 , we have f (x2 ) > f (x1 )

2
 y
√ √
y= 5x − 1 + 3x + 10
9
8
y=7
7
6
5
4
3
2
1
0 x
0 1 2 3 4 5 6 7 8 9

Problem 3 (4 marks).

Solve the equations

√ √ √
a) 10x + 19 + x + 1 = 3 −2x − 1;
√ √
b) 5x − 1 + 7x + 2 = 7;

1.3 Solving equations using substitution

Example 4. Solve the equation: r

x + 2 4(x − 1)
− =1
x−1 x+2

Solution: Let r
x+2
u= ,
x−1
then the equation becomes

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u− = 1 ⇒ u3 − u2 − 4 = 0, u 6= 0;
u2

(u − 2)(u2 + u + 2) = 0 ⇒ u = 2;

r
x+2 x+2
=2⇒ = 4 ⇒ x = 2.
x−1 x−1

Substituting into the original equation, we find this root to be valid.

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Answer: x = 2

Example 5. Solve the equation: √ √

3 3
2−x+ x+7=3

Solution: Let √ √
3 3
u= 2 − x; v = x+7
Then the equation becomes u + v = 3. We also notice that u3 + v 3 = 2 − x + x + 7 = 9 — so now we have
two equations for two variables, u and v. Their roots are u = 2, v = 1 or u = 1, v = 2.

Answer: x = 1, x = −6.

Problem 4 (6 marks). Solve the equations

r r
x+4 x−4
a) −2 = 73 ;
x−4 x+4
√
b) x2 + 3x2 − 6x + 7 = 7 + 2x;
√ √
c) 3 3x + 24 + 3 6x + 21 + 3 = 0.

2 Equation with modulus

The modulus (absolute value) function is defined as

"
x, x ≥ 0
|x| = (1)
−x, x < 0

Alternatively, we can say that √

|x| = x2 .

2.1 Equation |f (x)| = g(x)

The equation |f (x)| = g(x) can be written as two equations:


f (x) = g(x), f (x) ≥ 0
or


−f (x) = g(x), f (x) < 0.

If f (x) is easy to compute, you can solve the inequalities f (x) ≥ 0 and f (x) < 0 to identify extraneous
solutions. Otherwise it may be easier to simply find all solutions to both equations f (x) = g(x) and
f (x) = −g(x) and then eliminate the extraneous solutions by substituting the answers into the original
equations.

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Note that we used a square bracket in Eq. (1) to indicate that either the first or the second line must hold.
This is in contrast to curly brackets we use to mark systems of equations, in which all equations must hold
at the same time.

Example 6. Solve the equation:

x2 + 2x − 3|x + 1| + 3 = 0

Solution. We have two options:

" " "
x2 + 2x − 3(x + 1) + 3 = 0, x + 1 ≥ 0 x2 − x = 0, x ≥ −1 x = 1, x = 0
⇒ 2 ⇒
x2 + 2x − 3(−x − 1) + 3 = 0, x + 1 < 0 x + 5x + 6 = 0, x < −1 x = −2, x = −3

Luckily, all the roots satisfy the corresponding inequalities.

Answer: −3, −2, 0, 1.

Problem 5 (6 marks). Solve the equations:

a) |x + 2| + |x − 2| = 32x3 ;

b) |x2 − 4 × |x| + 4| = 4;
√ √
c) 4x2 − 4x + 1 + x2 + 2x + 1 = 2x + 1.

Problem 6 (4 marks). Solve the system of equations

(
y + |x + 1| = 1
|y − x| = 5.

2.2 Equation |f (x)| = |g(x)|


f (x) = g(x)
This equation is equivalent to or


f (x) = −g(x)

Problem 7 (6 marks). Solve the equations:

a) |x5 − 6x2 + 9x − 6| = |x5 − 2x3 + 6x2 − 13x + 6|;

p
b) |x2 + 14x + 47| − 1 = |x + 7| − 1.

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3 Equations with logarithms

3.1 Basic Equations

Sometimes one can solve equations by simplifying the expressions therein.

√
Example 7. (PAT 2012) If 5 = log2 16 + log10 0.01 + log3 x, determine x.

Solution: Express log3 x from the above equation as

√
log3 x = 5 − log2 16 − log10 0.01
= 5 − log2 2 − log10 10−1
4

= 5 − 4 + 1 = 2.

Therefore log3 x = 2, and the answer is x = 9.

Problem 8 (PAT 2015, 1 marks). Solve for x in the equation log2 x + log4 16 = 2.

Problem 9 (PAT 2018, 1 marks). Solve the following equation for x: logx 25 = log5 x.

3.2 Equation af (x) = bg(x) , a > 0, b > 0, b 6= 1.

Let us look first what happens if a = b, i.e. we are solving

af (x) = ag(x) . (2)

Taking the logarithm with base a of both sides we see that equation (2) is equivalent to2

f (x) = g(x).

Because the range of the exponential function is R, we are not introducing any extraneous solutions by doing
this simplification.

Example 8. Find all the values of x for which

22x+4 = 43x−2 .

Solution: Let us transform this equation as follows:

22x+4 = 22(3x−2)
2x+4
2 = 26x−4 .

Using Eq. (2) we see that solving the original equation is equivalent to solving 2x + 4 = 6x − 4, i.e. the only
value of x for which 22x+4 = 43x−2 is x = 2.
2 It is worth noting that this result is due to the exponential function being strictly monotonic (strictly monotonic functions

are always one-to-one). So if ax = ay , then necessarily x = y. This is not true for functions which are not strictly monotonic.
For example, sin x = sin y it does not mean that x = y

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Problem 10 (2 marks). Find all the values of x for which
2 √
2x −6x−2.5 = 16 2.

Now let us return to the original problem. How should we solve af (x) = bg(x) , a > 0, b > 0, b 6= 1?
As always in mathematics, it does help to transform your problem into something that you already know
how to solve. Let us write af (x) = bg(x) as

clogc a×f (x) = clogc b×g(x) , (3)

where c is some positive real number different from 1 (for example, e). Equation (3) we already know how
solve: it is equivalent to
logc a × f (x) = logc b × g(x).

Example 9. Find all the values of x for which

2
−1
7x = 2x−1 .

Solution: the equation in the question is equivalent to

ln 7 × (x2 − 1) = ln 2 × (x − 1),

which we can re-write as

ln 7 × (x − 1)(x + 1) = ln 2 × (x − 1).
We may be tempted to divide both sides by (x − 1), but before we do so, we should notice that the equation
holds when x − 1 = 0, so x = 1 is a solution. For x 6= 1, the division is safe, so
ln 2
ln 7 × (x + 1) = ln 2 ⇒ x = −1 + = −1 + log7 2 ≈ −0.644.
ln 7
Note that is is fully appropriate to just leave the answer in the form x = −1 + log7 2.

Problem 11 (3 marks). Find all values of x, for which

2 2 2 2
−1
2x · 5x − 0.5 · 2x · 32x − 5x + 32x = 0.

3.3 Other exponential equations

Sometimes a useful trick is to multiply an equation by a factor.

Example 10. Find all the values of x such that

9x + 6x = 22x+1 .

Solution: Let us write the above equation as

32x + 2x × 3x = 2 × 22x .

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Now divide both parts of the equation by 22x to get
 2x  x
3 3
+ − 2 = 0,
2 2

which is equivalent to  x   x 
3 3
+2 − 1 = 0. (4)
2 2
The first factor in the left-hand side of the above equation can never turn to zero, so it can be safely cancelled.
We obtain  x
3
− 1 = 0.
2
Therefore the answer is x = 0.

Problem 12. (3 marks) Solve the following equation:

4 × 25x + 5 × 16x = 9 × 20x .

Problem 13 (3 marks). Find all x for which

2 2
31+x + 31−x = 10.

3.4 Equation loga f (x) = loga g(x), a > 0, a 6= 1.

This equation can be transformed into

f (x) = g(x),
but we should remember that the logarithm log(·) is defined only for positive arguments. Hence our equation
is equivalent to the system

(
f (x) > 0
f (x) = g(x),
or (
g(x) > 0
f (x) = g(x).

Example 11. Find all x such that

x
log(x + 1) = − log .
2

Solution Let us rewrite our equation as

2
log(x + 1) = log .
x
and find x such that
2
x+1= ,
x

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which is equivalent to solving the quadratic equation

x2 + x − 2 = 0.

This equation has two roots: x = −2 and x = 1. However, only one of these, namely x = 1, satisfies the
condition x + 1 > 0

Answer: x = 1.

Problem 14 (6 marks). Solve the equations

1
a) log√x+1 9 + log3x 3 = 0;
4
√
b) log9 x + logx 3 − log9 x = 1.

4 Equations with a parameter

Problem 15 (4 marks). For which values of the parameter a does the equation 2 log(x + 5) − log(ax) = 0
have only one root?

Problem 16 (3 marks). For which values of the parameter a does the equation ||x − 1| − 2 + a| = 7 have
exactly 3 distinct roots?

5 Additional problems

Problem 17 (PAT 2006, 2 marks). Find all the values of x for which

(i) loge (e3x ) = 6;

(ii) log3 x2 = 2;

Problem 18 (PAT 2016, 2 marks). Given the two equations

 x
64
log4 = 13
16y

log10 10x + log3 3y = 1,

find x and y.

Problem 19 (PAT 2019, 2 marks). Solve the following equation for x:

ex + 9
= 2.
e−x + 5

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Problem 20 (4 marks). Solve the equation 3 × 4x−2 + 27 = a + a × 4x−2 and determine for which values
of the parameter a the equation has roots and for which it doesn’t.

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