Change of Logarithm Base

First, let's recall the definition of the log and understand what is the base of a logarithm:

The definition of the logarithm is:
$log_a⁡x=b$
$X=a^b$

Where:
$a$ is the base of the log
$X$ is what appears inside the log - which can also appear inside of parentheses
$b$ is the exponent to which we raise the base of the log to in order to obtain the number that appears inside of the log.

The rule states that if we raise base $a$ to the power of $b$ we obtain $X$.
In order to solve a log, we ask ourselves - to what power should we raise $a$ to obtain $X$.
The power $b$ that we found is the solution.

For example:
When we have the expression
$log_5⁡25=$

Let's ask ourselves –
To what power do we need to raise $5$ to in order to obtain $25$? The answer is $2$. Therefore:
$log_5⁡25=2$

Important to know - the log in the example is read as follows:
log base $5$ of $25$

You're probably asking yourself, why do we need to change the base of the logarithm?
Excellent question!
Sometimes in subtraction, addition, multiplication, or division exercises with different bases, it is easier for us to change the base of one of the logarithms so we can use the addition and subtraction formulas, and the same goes for multiplication and division.
So how do we do it? According to the following rule:

The Change of Base Rule for Logarithms:

$log_aX=\frac{log_{base~we~want~to~change~to}X}{log_{base~we~want~to~change~to}a}$

Let's look at an example:
Here is the exercise:
$log_8⁡16=$
Convert it to a log with base $2$.

Solution:
Upon observation this is an illogical solution... To what power should we raise $8$ to in order to obtain $16$?
Therefore, we can change the base of the log and solve the problem far more easily! Note that the instruction is to change the log to base $2$.
To change the base of the log, we'll use the formula:
In the numerator, we'll have log base $2$ (the base we want to change to) with the original log content $16$
And in the denominator, we'll have log base $2$ (the base we want to change to) with $8$ the original base.
We obtain the following:
$log_8⁡16=\frac{log_2⁡16}{log_2⁡8 }$
Are we done? Not yet. Now we must proceed to solve the logs until we obtain a number.
$log_2⁡16=4$
$log_28=3$
We insert the data into the exercise as seen below to obtain our solution:
$=\frac{4}{3}$
Amazing!

Now we'll increase the difficulty level and move on to another exercise with a variable where we'll also use the change of logarithm base rule:
$log_6⁡x+log_{36}⁡x=3$

Solution
First of all, don't panic. With the rule you just learned, you can solve this exercise very simply. Let's start step by step.

When encountering two logarithms with different bases, the first step is to convert the larger log to the smaller one because it will be simpler to solve.
Let's do it - we'll convert $log_{36}⁡x$ to base $6$ and get:
$log_{36}⁡x=\frac{log_6x}{log_6⁡36 }$

Insert the data into the exercise as follows:

$log_6⁡x+\frac{log_6⁡x}{log_6⁡36} =3$

Let's continue solving the exercise.
We can solve the denominator -
$log_6⁡36=3$
Insert into the exercise
$log_6⁡x+\frac{log_6⁡x}{2}=3〗$
Let's continue solving the problem and write the exercise in a simpler form:

$log_6⁡x+0.5 log_6⁡x=3$
Let's combine like terms:
$1.5 log_6x=3$

$log_6 x=2$
Solve the following –
$x=6^2$
$x=36$

Tip - Sometimes it's useful to substitute an auxiliary variable $t$ instead of the entire log.
When reached this stage in the exercise -
$log_6⁡x+\frac{log_6⁡x}{2}=3$
We can substitute
$log_6 x=t$
as follows:
$t+\frac{t}{2}=3$
$1.5t=3$
$t=2$

But notice, this is not the result. Now we need to substitute $2$ instead of $t$
and determine $x$
we obtain the following solution :
$log_6 x=2$
$6^2=x$
$x=36$