Clash of logs

$log_{10^{2}} (10^{2} \times 10^{2})$

$= log_{10^{2}} 10^{2} + log_{10^{2}} 10^{2}$

$= 1 + 1 = 2$

We know that

$log_{a^b}m$ = $\dfrac{1}{b}$ $log_{a} m$

And

$log_{a}(m^b)$ = $b. log_{a}m$

Using this property

$log_{10^2}(10^4)$ = $\dfrac{2}{2}$ $log_{10}10$ = $\boxed{2}$

What I do when answering logarithms such as $\log_{a}b$ is to ask myself: What will I raise to $a$ to get $b$ ? In this case, $(10^2)^x = 10^4$ $(10^2)^x=(10^2)^2$ $x=2$

$log_{10^2}10^4 = \dfrac{1}{2}log_{10}10^4 = 4 \times \dfrac{1}{2} log_{10}10 = 2 \times 1 = \color{#D61F06}{\boxed2}$

$log_{a^n}$ b^m)=[m/n][\(log_{a} (b)

(log 10^4) / (log 10^2) =(4 (log10)) / (2 (log10)) [its log rule that power always comes before log. e.g. log 10^4 = 4*log10] Now, log 10 gets divided by log 10. And 4/2= 2.