What is the proof for log(a^c)=clog(a)?

Question

What is the proof for log(a^c)=clog(a)?

Answer 1

Well, if you'll recall that logarithms are really nothing but exponents, it's fairly easy to see.  Since you chose log, let's use base 10.

Let's see if:

10^(log(a^c)) = 10^(clog(a)).

Let's start with the first expression, 10^(log(a^c).

We know that 10^(log(b)) = b, so

10^(log(a^c)) = a^c.

10^(clog(a)) = (10^c)^(log(a)) = 10^(log(a))^c = a^c.

So, the expressions yield the same value, therefore they're equal.

Hope this helps.

Answer 2

log(a^1) ≡ log(a) ≡ 1log(a)

Assume log(a^k) = klog(a)

Then log(a^(k+1)) = log(a*a^k) = log(a^k) + log(a) =

klog(a) + log(a) = (k + 1)log(a)

so

log(a^c) = clog(a)