One way is to manipulate both sides into having the same base, and then equating exponents.
\(\displaystyle \mbox{ 4^{2x} = \left(2^2\right)^{2x} = 2^{4x}}\)
\(\displaystyle \mbox{ }\)and
\(\displaystyle \mbox{ 8^8 = \left(2^3\right)^{8} = 2^{24}}\)
So the equation becomes
\(\displaystyle \mbox{ 2^{4x} = 2^{24}}\)
We now have equal bases, so can equate components:
\(\displaystyle \mbox{ 4x = 24}\)
\(\displaystyle \mbox{ \Rightarrow x = 6}\)
Alternatively, from the start, you could take logs (of any consistent base) of both sides.
