Subtraction of Logarithms

First, let's recall what is the definition of $log$?
$log_a⁡x=b$
Where $a$ is the base of the log (usually $10$)
$b$ is the exponent we raise $a$ to
$X$ is the number that appears inside the log, sometimes in parentheses, and it's the number that we get when $a$ is raised to the power of $b$
meaning:
$X=a^b$
For example, if we have an exercise like this:
$log_5⁡25=$
Determine which power we need to raise $5$ to in order to obtain$25$....?
The answer is to the power of $2$ and therefore the solution is $2$.

To easily subtract logarithms with the same base, all you need to know is the following rule about subtracting logarithms with identical bases:
$log_a⁡x-log_a⁡y=log_a⁡\frac{x}{y}$
The rule states that if you want to subtract $2$ logs with the same base, you can write them as $1$ log and divide the numbers inside the log. This will sometimes make it easier to solve.
Note - The numerator will always be the first log from which we subtract, and the denominator will always be the second log that we subtract from the original.

Let's look at an example:
$log_7⁡147-log_7⁡3=$
At first glance, this problem looks intimidating. However you'll soon see how using the subtraction rule for identical logarithms makes it simple.
Which power should we raise $7$ to in order to obtain $147$?... Which power should we raise $7$ to in order to obtain $3$?
All we need to do is divide the numbers that appear in the logarithm while keeping the base the same - $7$.
Of course, we'll do this in order - in the numerator we'll put the first number $147$ and in the denominator we'll put the second number $3$.
We obtain the following:
$log_7⁡147-log_7⁡3=log_7⁡\frac{147}{3}$
As well as:
$log_7⁡\frac{147}{3}=log_7⁡(49)$
Now it's much easier for us to solve the equation!
We know that we need to raise $7$ to the power of $2$ to obtain $49$ and therefore the entire answer to this problem is $2$.
$log_7⁡(49)=2$

Note - this rule is only valid in cases where the base is identical. If the base was not the same in both logarithms, we could not use this rule.

In order to subtract logs with different bases, you need to use the log base change rule.
The goal is - to convert both logs to the same base.

Meet the change of base formula for logs.
$log_aX=\frac{log_{the~base~we~want~to~change~to}X}{log_{the~base~we~want~to~change~to}a}$

And now for the explanation:
When we have a log with base $a$ and we want to convert it to another log, we will always convert it to a fraction in the following way:

1. Draw a fraction line.
2. In the numerator, write the logarithm with the desired new base as well as what was in the original logarithm.
3. In the denominator, write the logarithm with the desired new base where the inside of the logarithm will be the base of the original logarithm.

Let's look at an example:

$log_{25⁡}2=$
Convert the following logarithm to base $5$:

$log_{25}625=\frac{log_5⁡625}{log_5⁡25}$

In the numerator we'll write the log with base $5$, the base we want to convert to. The number inside the log in the numerator will be the original number that appears inside the log - which is $625$.

In the denominator, we will write again log base $2$, the base we want to convert to, but this time, the number inside the log will be the original base - which is $25$
Now we can solve the problem easily. We will obtain the following answer:

$\frac{log_5⁡625}{log_5⁡25} =\frac{4}{2}=2$

Advanced exercise:
Now you can solve logarithm subtraction with different bases:
$log_3⁡x-log_9⁡x=2$
We want to convert both logarithms to the same base, and usually we choose the smaller base - $3$.
Therefore:
$log_9⁡x=\frac{log_3⁡x}{log_3⁡9}$
Let's now rewrite the exercise and insert the data that we obtained:
$log_3⁡x-\frac{log_3⁡x}{log_3⁡9} =2$

Let's insert $log_3⁡9=2$
and obtain the following:
$log_3⁡x-\frac{log_3⁡x}{2}=2$
$log_3⁡x-0.5log_3⁡x=2$

$0.5 log_3⁡x=2$
$log_3⁡x=4$
$x=3^4$

$x=81$