Is $\log (t^2 (l/c)) = \log (t^2) \log (l/c)$?

Question

I'm new in this forum want to ask a beginner question about logarithm:

Is $\log (t^2 (l/c)) = \log (t^2) \log (l/c)$?

Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you write what your thoughts are on the problem and include your efforts (work in progress) in this and future posts and in what context you have encountered the problem; this will prevent people from telling you things you already know, and help them give their answers at the right level. — JKnecht, Feb 10 '16 at 00:27
If you're learning logarithms, you should have been introduced to the product rule, quotient rule, and power rule. Note that in the log argument you have a product...what can you say about that? — Eleven-Eleven, Feb 10 '16 at 00:29
In general this is what you have: $\log(ab) = \log(a)+\log(b)$. Instead of multiplication, you need addition up there. — Cameron Williams, Feb 10 '16 at 00:31
@CameronWilliams many thanks i didn't use logarithm for a long time so i know that it's beginner question i searched but i though i searhed with wrong way , Thanks agian — user5520049, Feb 10 '16 at 00:32

Eleven-Eleven · Answer 1 · 2016-02-10T00:41:45.183

There are three main rules:

1) Product rule:

$$\log_b{xy}=\log_b{x}+\log_b{y}$$

2) Power rule: (can be derived from (1))

$$\log_b{x^n}=n\cdot \log_b{x}$$

2) Quotient rule: (can be derived from (1) and (2))

$$\log_b{\frac{x}{y}}=\log_b{x}-\log_b{y}$$

Now note that the expression in your argument is

$$t^2\cdot\frac{l}{c}$$

which involves all three rule cases. See if you can work slowly rule by rule and unravel the expression.

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