Changing the Base of a Logarithm

First, let's recall the definition of the log:

$log_a⁡x=b$

When
$a$ is the base of the log (note that in the calculator the base is $10$ by default)
$b$ is the exponent to which we raise $a$
$X$ is the number we obtain when $a$ is raised to the power of $b$, also called the number inside the log. Sometimes it appears in parentheses.
In other words:
$X=a^b$

For example, if we encounter an exercise like this:
$log_3⁡27=$
Let's consider what power we need to raise $3$ to in order to obtain$27$....?
The answer is to the power of $3$ and therefore the solution is $3$.
If we encounter an exercise like this with a variable in the log base:
$log_x⁡36=2$
We'll calculate as follows
$x^2=36$
$x=6$

If we encounter an equation like this with a variable inside the log:
$log_2⁡x=5$
We calculate as follows:
$2^5=x$
$x=32$
Great! After reviewing the definition of log, we'll proceed to change the base of logarithms.

Inverting the log base is a situation where we want to switch the positions of the log base and the log content. To do this, we will use the following formula:
$log_a⁡x=\frac{1}{log_x⁡a}$

Let's look at an example:
Solve the logarithm
$log_{25}⁡5=$

Solution:
We can simply use the formula to solve the problem. In the numerator we'll write $1$
and in the denominator we will the inverse logarithm
We'll obtain the following :
$log_{25}⁡5=\frac{1}{log_5⁡25 }$
Now we can solve the problem as shown below
$log_5⁡25=2$
We'll insert the data into the exercise as follows to obtain our solution:
$log_{25}⁡5=\frac{1}{2}$

When you need to calculate a logarithm in a specific base, but your calculator only works with base $10$, use the change of base formula! In fact, you can use it for addition, subtraction, multiplication, or division problems with different bases!
Instead of racking your brain looking for a solution, just use the formula and see how easily you can solve the problem !

Let's practice!

Solve the exercise:
$log_4⁡64+log_{16} 4=$

Solution
The first thing we'll do is simply look at the exercise. At first glance, we can see that the logarithms are inverse. This means - if we perform a base change for one of the logarithms in the exercise, we can easily solve it using the addition formula for logarithms with identical bases.

Tip - It's better to convert the larger base to the smaller base.
$log_{16}4=\frac{1}{log_416}$
According to the formula we learned. Now let's insert the data we obtained in the exercise as follows:
$log_4⁡64+\frac{1}{log_416}=$
Let's continue solving the problem. Note there's no need to find a common denominator. Just solve the logs as they are.
$log_4⁡64=3$
$\frac{1}{log_416}=\frac{1}{2}$
Let's substitute the data into the exercise as follows:
$3+\frac{1}{2}=3.5$

Note - There are two ways to change the logarithm base.
One way is using the formula you learned here - which actually swaps the log base and the log content.
The second way is to use the change of base formula.
Here it is -
$log_aX=\frac{log_{the~base~you~want~to~change~to}X}{log_{the~base~you~want~to~change~to}a}$

Let's use this formula to convert $log_{16} 4$ to a logarithm with a base $4$.
In the numerator we'll have a logarithm with a base $4$ - the base we want to convert to, and the content will be $4$ the original content
In the denominator we'll have a logarithm with base $4$ - the base we want to convert to, and the content will be $16$ = the original base

We obtain the following:
$log_{16} 4=\frac{log_44}{log_416}$
Let's continue:
$log_{16} 4=\frac{1}{2}$
Notice - we obtain $1 \over 2$
exactly like in the first formula.

Now we'll solve an advanced exercise that combines both methods together!
Don't worry, we'll solve it step by step.

Solve the exercise –
$log_64⁡2+log_8⁡16=$

Solution:
The first thing we'll do is change the first logarithm with the base $64$ given that the log with base $2$ is easier to solve. We obtain the following:
$log_{64}⁡2=\frac{1}{log_2⁡64 }$
Let's substitute as follows:
$\frac{1}{log_2⁡64} +log_8⁡16=$
Now we'll use the second formula you learned - changing the logarithm base and converting the log 8 to log $2$.
We obtain the following:
$log_8⁡16=\frac{log_2⁡16}{log_2⁡8 }$
Let's insert the data as follows:
$\frac{1}{log_2⁡64} +\frac{log_2⁡16}{log_2⁡8} =$

Now we'll solve the logs as they are and obtain the following solution:
$\frac{1}{6}+\frac{4}{3}=1.5$