A question about simplification of an expression involving log: $16777216^\frac{\log(64n)}{3\log(4)}$

Question

I have a practice question regarding simplification of the following expression:

$16777216^\frac{\log(64n)}{3\log(4)}$

So I have tried to do this:

$(64^4)^\frac{\log64n}{\log64}$

and now I got stuck. Maybe there are something more I could do to the exponent term using change of base. Say I simplify it further to: $(64^4)^{\log_{64}{(64n)}}$

But this is as far as I can go, I think there might be more simplification that could be done. But not sure how to proceed further. Could someone help a bit?

Answer 1

Just switch the exponents around as

$$(a^b)^c = a^{bc} = (a^c)^b \implies \left( 64^4 \right)^{\log_{64}(64n)} = \left( 64^{\log_{64}(64n)} \right)^4=(64n)^4 = 16,777,216 \cdot n^4$$

Answer 2

$(64^4)^{\log_{64}64n} = (64^4)^{1+\log_{64}n} = 64^4 \times (64^4)^{\log_{64}n} = 64^4 \times (64^{\log_{64}n})^4 = 64^4n^4$