Solve the Equation:

(log[base2]({2^2x}+1)-x)(3log[base8]({2^-2x}+1)+x-2)=15

I know that I want to multiple the left hand side to get the form
log[base something](equation)=15
so that [base something]^15=equation but I do not know how to multiple the equation with the different bases.

Help...

(log[base2]({2^2x}+1)-x)(3log[base8]({2^-2x}+1)+x-2)=15
just to restate the problem:

\left(\log_2\left(2^{2x}+1\right)-x\right)\left(3\log_8\left(2^{-2x}+1\right)+x-2\right)=15

I am pretty sure that this won't be trivial to solve.
Your question: what is \log_a x*\log_b y? This doesn't really reduce into something else.

I'll look at this problem a bit more but I suspect that it isn't really solvable algebraically.

maybe try to look at the property:
\log_ab=\frac {\log_cb} {\log_ca}
so:
\log_8x=\frac {\log_2x} {\log_28}=\frac {\log_2x} 3

it seems to me that we might be able to go somewhere with that. The 3 that pops out looks awfully suspicious :)

is this the whole problem or is this part of a larger problem? It doesn't seem like the equation you've given is solvable.

This is the whole problem.

I would use tech to solve this monster.

I ran it through my TI and got x=2.56920814769...