Multiplication of Logarithms

In order to solve perform the multiplication of logarithms, you need to know the following rule:
$log_a⁡(x\cdot y)=log_a⁡x+log_a⁡y$
When there is multiplication inside the log, we can split the log into an addition problem with the same base where one log will have the content of one of the factors and the second log will have the content of the second factor.
You can convert the multiplication problem to an addition problem and an addition problem to a multiplication problem with one log according to the rule as long as the base is the same.

Let's look at an example:
$log_4⁡(64\cdot16)=$
We have an expression with a logarithm containing a multiplication operation. If we were to multiply what's inside of the parentheses, we would obtain something like this:
$log_4⁡(1024)=$

Of course, thanks to the rule, we can solve this exercise in a much easier and faster way.
We just need to split the exercise into an addition exercise with $2$ identical bases - in this exercise the base is $4$.
It should look like this:
$log_4⁡(64\cdot16)=log_4⁡64+log_4⁡16$

Now we can solve the exercise with greater ease!

Reminder -
The definition of a logarithm is:
$log_a⁡x=b$
$X=a^b$
Where:
$a$ is the base of the logarithm
$X$ is what appears inside the logarithm. It can also appear in parentheses.
$b$ is the exponent to which we raise the base of the logarithm to obtain the number inside the logarithm.

And so we ask ourselves - to what power do we need to raise $4$ to in order to obtain $64$? The answer is $3$.
And to what power do we need to raise $4$ to in order to obtain $16$? The answer is $2$.
We obtained the following:
$log_4⁡64+log_4⁡16=3+2=5$

Let's solve another exercise!
$log_6⁡(36\cdot216)=$

Solution:
To solve this exercise, we'll use the rule we learned about multiplying logarithms.
We can split the exercise into an addition exercise where the base is identical and equals $6$.
As follows:
$log_6⁡36+log_6⁡216=$
Now we can solve the problem more easily.
We know that in order to obtain $36$ we need to raise $6$ to the power of $2$ therefore
$log_6⁡36=2$
and in order to obtain $216$ we need to raise $6$ to the power of $3$ therefore
$log_6⁡216=3$
Let's substitute the data into the exercise as follows:
$2+3=5$
$5$ is the final answer.

Let's proceed to another exercise!
\(log_6⁡2+log_6⁡18=\

Pay attention! At first glance, this exercise appears to be an addition of logarithms with the same base... however! We are using the rule we learned about multiplying logarithms!

Solution Method:
Let's first try to solve the exercise without the rule -
$log_6⁡2=$
If we think about which power we need to raise $6$ to in order to obtain $2$... we encounter a problem. This is not an intuitive solution
The again encounter the same problem for $log_6⁡18=$
Which power do we need to raise $6$ to in order to obtain $18$? Also a good question..
Therefore, we use the rule that states we can multiply the $2$ log contents with the same base. As seen below:

$log_6⁡2+log_6⁡18=log_6⁡(2\cdot18)$
$=log_6⁡36$
How wonderful! We now understand that in order to reach $36$ we need to raise $6$ to the power of $2$!
Therefore the solution is $2$. This is the final answer.

What did we learn? The multiplication rule we learned follows the commutative property.
It works on both sides - you can convert a multiplication expression into an addition expression, and an addition expression can be converted into a single log with multiplied content.
However - only if the base is identical.

Note - If there is multiplication of logarithms with different bases, you can try to convert the log base according to the rules you learned.