Logarithm Question, Simplifying

Simplify the following:
2log[5]x^2 ÷ 1/3log[5]x

I have no idea how it becomes 12.

I've tried:
log[5]x^4 ÷ log[5]x^1/3
log[5]x^(4-1/3)
log[5]x^(11/3)

Idk. Help please.

Your logarithmic identites are wrong.

log(AB) = log(A) + log(B)
log(A/B) = log(A) - log(B)


But instead, you used...
log(A) / log(B) ≠ log(A - B) which is a big no no!

So... (assume the logs are base 5)

2log(x^2) ÷ (1/3)log(x)
= 2(2log(x)) ÷ (1/3)log(x)
= better to see this as a fraction rather than a division problem...
= 4log(x) / (1/3)log(x)

The logx's on both denominator and numerator cancel out
= 4 / (1/3)
= 4 * 3
= 12

Hope this helps :D

When you're dealing with very complicated exponents within the logs, just take a step back and put all the exponents on the front.

there is something wrong with the question though. Did you mean to put a subtraction sign rather than division? because the minus sign means you divide...

log(5)x^4
--------------
1/3log(5)x
=4log(5)x
----------
1/3log(5)x
=4.3 log(5)x/log(5)x
=12