log2 32 – log4 16 + log3 27
An explanation will help too 😀
3 Answers

There are 2 ways to do this particular problem. The main purpose of this exercise is to teach you the following rule of logarithms:
log[a](a^b) = b
Now, let’s look at your problem:
log[2](32) – log[4](16) + log[3](27)
Let’s rewrite this a bit:
32 = 2^5
16 = 4^2
27 = 3^3
log[2](2^5) – log[4](4^2) + log[3](3^3) =>
5 * log[2](2) – 2 * log[4](4) + 3 * log[3](3) =>
5 * 1 – 2 * 1 + 3 * 1 =>
5 – 2 + 3 =>
6
Now, what’s the second way? Well, that involves something known as the “Change of Base” formula
log[a](b) =>
log[c](b) / log[c](a)
For instance, log[2](32) can be rewritten as log[10](32) / log[10](2)
log[2](32) – log[4](16) + log[3](27) =>
log(32) / log(2) – log(16) / log(4) + log(27) / log(3) =>
log(2^5) / log(2) – log(4^2) / log(4) + log(3^3) / log(3) =>
5 * log(2) / log(2) – 2 * log(4) / log(4) + 3 * log(3) / log(3) =>
5 – 2 + 3 =>
6
We could also try and get a common denominator and solve that way, but I have a sick feeling in my gut that tells me that that particular method would be monstrous to behold.

Subtract Logarithms

log2 32 – log4 16 + log3 27
= log2 2^5 – log4 4^2 + log3 3^3
= 5 – 4 + 3
= 4
The explanation is log_b(b^x) = x log_b(b) = x*1 = x