Logarithmic
Functions
The logarithmic function to the base a, where a > 0 and a 1
is defined:
y = logax if and only if x = a y
logarithmic
form exponential
form
When you convert an exponential to log form, notice that the
exponent in the exponential becomes what the log is equal to.
Convert to log form: 16 4 2 log 416 2
Convert to exponential form:
31
1
log 2 3 2
8 8
LOGS = EXPONENTS
With this in mind, we can answer questions about the log:
This is asking for an exponent. What
log 2 16 4 exponent do you put on the base of 2 to
get 16? (2 to the what is 16?)
1 What exponent do you put on the base of
log 3 2 3 to get 1/9? (hint: think negative)
9
log 4 1 0
What exponent do you put on the base of
4 to get 1?
1 When working with logs, re-write any
1
log33 33
log 2 radicals as rational exponents.
What exponent do you put on the base of
2 3 to get 3 to the 1/2? (hint: think rational)
Example 1
Solve for x: log 6 x 2
Solution:
Let’s rewrite the problem in
exponential form.
62 x
We’re finished !
Example 2
1
Solve for y: log5 y
25
Solution: Rewrite the problem in
exponential form.
1
5
y 1
25
Since 5 2
25
5y 5 2
y 2
Example 3
Evaluate log3 27.
Solution:
Try setting this up like this:
log3 27 y Now rewrite in exponential form.
3y 27
3y 33
y3
Example 4
2
Evaluate: log7 7
Solution:
log7 7 y
2
First, we write the problem with a variable.
7y 72 Now take it out of the logarithmic form
and write it in exponential form.
y2
Example 5
log 4 16
Evaluate: 4
Solution:
4 log 4 16
y First, we write the problem with a variable.
log4 y log4 16 Now take it out of the exponential form
and write it in logarithmic form.
Just like 23 8 converts to log2 8 3
y 16
Finally, we want to take a look at the Property of Equality
for Logarithmic Functions.
Suppose b 0 and b 1.
Then logb x1 logb x 2 if and only if x1 x 2
Basically, with logarithmic functions,
if the bases match on both sides of the
equal sign , then simply set the
arguments equal.
Example 1
Solve: log3 (4x 10) log3 (x 1)
Solution:
Since the bases are both ‘3’ we simply set the
arguments equal.
4x 10 x 1
3x 10 1
3x 9
x 3
Example 2
Solve: log8 (x 14) log8 (5x)
2
Solution:
Since the bases are both ‘8’ we simply set the arguments equal.
x 14 5x
2
x 2 5x 14 0 Factor
(x 7)(x 2) 0
(x 7) 0 or (x 2) 0
x 7 or x 2 continued on the next page
Example 2
continued
Solve: log8 (x 14) log8 (5x)
2
Solution:
x 7 or x 2
It appears that we have 2 solutions here.
If we take a closer look at the definition of a
logarithm however, we will see that not only
must we use positive bases, but also we see
that the arguments must be positive as well.
Therefore -2 is not a solution.
Let’s end this lesson by taking a closer look at
this.
Our final concern then is to
determine why logarithms like the
one below are undefined.
log 2 (8)
Can anyone give
us an explanation ?
log 2 (8) undefined WHY?
One easy explanation is to simply rewrite this
logarithm in exponential form.
We’ll then see why a negative value is not
permitted.
log 2 (8) y First, we write the problem with a variable.
2 8
y
Now take it out of the logarithmic form
and write it in exponential form.
What power of 2 would gives us -8 ?
1
2 8 and 2
3 3
8
Hence expressions of this type are undefined.
Characteristics about the Characteristics about the
Graph of an Exponential Graph of a Log Function
Function f x a x a > 1 f x log a x where a > 1
1. Domain is all real numbers 1. Range is all real numbers
2. Range is positive real 2. Domain is positive real
numbers numbers
3. There are no x intercepts
3. There are no y intercepts
because there is no x value
that you can put in the
function to make it = 0
4. The y intercept is always 4. The x intercept is always
(0,1) because a 0 = 1 (1,0) (x’s and y’s trade places)
5. The graph is always 5. The graph is always
increasing increasing
6. The x-axis (where y = 0) is 6. The y-axis (where x = 0) is
a horizontal asymptote for a vertical asymptote
x-
Exponential Graph Logarithmic Graph
Graphs of
inverse
functions are
reflected about
the line y = x
Transformation of functions apply
to log functions just like they apply
to all other functions so let’s try a
couple.
up 2 f x log10 x
f x 2 log10 x
Reflect about x axis
f x log10 x
left 1
f x log10 x 1
Remember our natural base “e”?
We can use that base on a log.
log e 2.7182828 1 What exponent do you put
on e to get 2.7182828?
ln
Since the log with this base occurs
ln 2.7182828 1 in nature frequently, it is called the
natural log and is abbreviated ln.
Your calculator knows how to find natural logs. Locate
the ln button on your calculator. Notice that it is the
same key that has ex above it. The calculator lists
functions and inverses using the same key but one of
them needing the 2nd (or inv) button.
Another commonly used base is base 10.
A log to this base is called a common log.
Since it is common, if we don't write in the base on a log
it is understood to be base 10.
log 100 2 What exponent do you put
on 10 to get 100?
1
log 3 What exponent do you put
on 10 to get 1/1000?
1000
This common log is used for things like the richter
scale for earthquakes and decibles for sound.
Your calculator knows how to find common logs.
Locate the log button on your calculator. Notice that it
is the same key that has 10x above it. Again, the
calculator lists functions and inverses using the same
key but one of them needing the 2nd (or inv) button.
Another commonly used base is base 10.
A log to this base is called a common log.
Since it is common, if we don't write in the base on a log
it is understood to be base 10.
Log10 100
log 100 2 What exponent do you put
on 10 to get 100?
1
log 3 What exponent do you put
on 10 to get 1/1000?
1000
This common log is used for things like the richter
scale for earthquakes and decibles for sound.
Your calculator knows how to find common logs.
Locate the log button on your calculator. Notice that it
is the same key that has 10x above it. Again, the
calculator lists functions and inverses using the same
key but one of them needing the 2nd (or inv) button.
The secret to solving log equations is to re-write the
log equation in exponential form and then solve.
log 2 2x 1 3 Convert this to exponential form
check:
2 2x 1
3
7
log 2 2 1 3
8 2x 1 2
7 2x log 2 8 3
7
x This is true since 23 = 8
2
Logarithmic Functions
Consider f x a x
where a>0 and a≠1
This is a one-to-one function, therefore it has an inverse.
The inverse is called the logarithmic function with base a.
log a x y a x y
Example: 16 24 4 log 2 16
The most commonly used bases for logs are 10: log10 x log x
and e: log e x ln x
y ln x is called the natural logarithm function.
Properties of Logarithms
a log a x
x log a a x x a 0 , a 1 , x 0
Since logs and exponentiation are inverse functions, they
“un-do” each other.
Product rule: log a xy log a x log a y
x
Quotient rule: log a log a x log a y
y
Power rule: log a x y log a x
y
ln x
Change of base formula: log a x
ln a
Properties of Logarithms
a log a x
x log a a x x a 0 , a 1 , x 0
Since logs and exponentiation are inverse functions, they
“un-do” each other.
Product rule: log a xy log a x log a y
x
Quotient rule: log a log a x log a y
y
Power rule: log a x y log a x
y
ln x
Change of base formula: log a x
ln a
Product rule
The logarithm product rule is a fundamental property of
logarithms that states:
log(a × b) = log(a) + log(b)
This rule allows us to simplify complex logarithmic
expressions by breaking them down into smaller,
more manageable parts.
Product rule
Problem
Simplify the expression: log(4 × 9)
Solution
Using the logarithm product rule, we can rewrite the
expression as:
log(4 × 9) = log(4) + log(9)
Next, we can simplify the individual logarithms:
log(4) = log(2^2) = 2log(2)
log(9) = log(3^2) = 2log(3)
Product rule
Now, we can substitute these simplified expressions
back into the original equation:
log(4 × 9) = 2log(2) + 2log(3)
Finally, we can combine the logarithms using the
logarithm product rule:
log(4 × 9) = log(2^2 × 3^2) = log(36)
Therefore, the simplified expression is:
log(4 × 9) = log(36)
Product rule
Simplify the expression: log(8 × 27)
Solution
Using the logarithm product rule, we can rewrite the
expression as:
log(8 × 27) = log(8) + log(27)
Next, we can simplify the individual logarithms:
log(8) = log(23) = 3log(2)
log(27) = log(33) = 3log(3)
Product rule
Now, we can substitute these simplified expressions
back into the original equation:
log(8 × 27) = 3log(2) + 3log(3)
Finally, we can combine the logarithms using the
logarithm product rule:
log(8 × 27) = log(23 × 33) = log(216)
Therefore, the simplified expression is:
log(8 × 27) = log(216)
Quotient rule
The logarithm quotient rule is a fundamental property
of logarithms that states:
log(a / b) = log(a) - log(b)
This rule allows us to simplify complex logarithmic
expressions by breaking them down into smaller,
more manageable parts.
Quotient rule
Problem
Simplify the expression: log(12 / 4)
Solution
Using the logarithm quotient rule, we can rewrite the
expression as:
log(12 / 4) = log(12) - log(4)
Next, we can simplify the individual logarithms:
log(12) = log(2^2 × 3) = 2log(2) + log(3)
log(4) = log(2^2) = 2log(2)
Quotient rule
Now, we can substitute these simplified expressions
back into the original equation:
log(12 / 4) = (2log(2) + log(3)) - 2log(2)
Simplifying further, we get:
log(12 / 4) = log(3)
Therefore, the simplified expression is:
log(12 / 4) = log(3)
Quotient rule
Simplify the expression: log(48 / 6)
Solution
Using the logarithm quotient rule, we can rewrite the
expression as:
log(48 / 6) = log(48) - log(6)
Next, we can simplify the individual logarithms:
log(48) = log(24 × 3) = 4log(2) + log(3)
log(6) = log(2 × 3) = log(2) + log(3)
Quotient rule
Now, we can substitute these simplified expressions
back into the original equation:
log(48 / 6) = (4log(2) + log(3)) - (log(2) + log(3))
Simplifying further, we get:
log(48 / 6) = 3log(2)
Therefore, the simplified expression is:
log(48 / 6) = 3log(2)
