solve:
$\displaystyle 2^{2x} = 5^x$
i went this far:
$\displaystyle \log{2^{2x}} = \log{5^x}$
$\displaystyle 2x\log{2} = x\log{5}$
how would you finish it?
no you would not. Krizalid is right, logs are definitely not needed. however, if you do want to force the issue...
we have $\displaystyle 4^x = 5^x$
take the log of both sides
$\displaystyle \Rightarrow \log 4^x = \log 5^x$
$\displaystyle \Rightarrow x \log 4 = x \log 5$
$\displaystyle \Rightarrow x \log 4 - x \log 5 = 0$
$\displaystyle \Rightarrow x ( \log 4 - \log 5) = 0$
$\displaystyle \Rightarrow x = 0 \mbox{ or } \log 4 - \log 5 = 0$
this second option is an impossibility, thus $\displaystyle x = 0$ as we got from Krizalid's much simpler solution