Addition of Logarithms

First, let's recall what is the definition of $log$?
$log_a⁡x=b$
When $a$ is the base of the log (usually $10$)
$b$ is the exponent to which we raise $a$
$X$ which sometimes appears in parentheses, is the number we obtain when $a$ is raised to the power of $b$, also called the number inside the log.
In other words:
$X=a^b$

For example, if we encounter an exercise like this:
$log_6⁡36=$

Determine which power we need to raise $6$ by in order to obtain $36$....?
The answer is the power of $2$ and therefore the solution is $2$.

In order to easily add logarithms with the same base, all you need to know is the following rule:
$log_a⁡x+log_a⁡y=log_a⁡(x\cdot y)$
The rule states that if you want to add $2$ logs with the same base, you can write them as $1$ log and multiply the numbers inside the log. This will often simplify the solution process.

Let's look at an example:
$log_8⁡32+log_8⁡2=$
If you didn't know the above, you would almost certainly encounter a problem.
To which power should we raise $8$ by in order to obtain $2$?... And to which power should we raise $8$ by in order to obtain $32$?
This is where the rule that you learned above comes in handy!
All you need to do is multiply the numbers that appear in the log whilst maintaining the same base– $8$.
Thus we obtain the following:
$log_8⁡32+log_8⁡2=log_8⁡(32\cdot2)$
As well as:
$log_8⁡(32\cdot2)=log_8⁡(64)$
Now it's much easier for us to solve the equation!
We know that we need to raise $8$ to the power of $2$ in order to obtain $64$ and therefore the entire answer to this exercise is $2$.
$log_8⁡(64)=2$

Note - This rule is valid only in cases where the base is identical. If the base was not the same in both logarithms, we could not use this rule.
Remember!
When you have the same base in both logarithms and an addition operation between them, you can multiply the numbers inside the logarithms and keep the base as it is - addition~multiplication

What happens when there is an addition exercise with logarithms of different bases?
In order to add logarithms with different bases, it's important to know the rule that allows us to change the base of a logarithm.
The goal is - to convert both logarithms to the same base.

Meet the logarithm change of base rule:
$log_aX=\frac{log_{the~base~we~want~to~change~to}X}{log_{the~base~we~want~to~change~to}a}$

Now for the explanation:
When we have a logarithm with base $a$ for example and we want to convert it to another logarithm:

1. Draw a fraction line.
2. In the numerator, write the logarithm with the desired new base and what was in the original logarithm.
3. In the denominator, write the logarithm with the desired new base and inside it put the base of the original logarithm.

Let's look at an example:

$log_8⁡16=$
Convert the following logarithm to base $2$:

$log_8⁡16=\frac{log_26}{log_28}$

In the numerator, we'll write log base $2$, which is 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 $16$.

In the denominator, we'll 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 $8$
Now we are able to solve the problem easily. We obtain the following:

$\frac{log_2⁡16}{log_2⁡8} =\frac{4}{3}$

Remember - whenever you want to convert to a different log base, you'll need to convert the log to a fraction according to the rule you just learned.
Advanced exercise: Now you can solve addition of logarithms with different bases:
$log_{25}⁡x+ log_5⁡x=3$
We want to convert both logs to the same base and usually we'll choose the smaller base - $5$.
Therefore:

$log_{25⁡}x=\frac{log_5⁡x}{log_5⁡25}$
Let's now rewrite the exercise and insert our data:

$\frac{log_5⁡x}{log_5⁡25} ⁡+ log_5⁡x=3$

Let's insert $log_5⁡25=2$
and obtain the following:
$\frac{log_5⁡x}{2}⁡+ log_5⁡x=3$
$0.5⁡log_5⁡x+log_5⁡x=3$

$1.5 log_5⁡x=3$
$log_5⁡x=2$
$x=5^2$

We obtained the solution:
$x=25$