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helpless06 · Nov 9, 2007, 09:37 AM

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...

asterisk_man · Nov 9, 2007, 11:28 AM

(log[base2]({2^2x}+1)-x)(3log[base8]({2^-2x}+1)+x-2)=15
just to restate the problem:
$<br /> \left(\log_2\left(2^{2x}+1\right)-x\right)\left(3\log_8\left(2^{-2x}+1\right)+x-2\right)=15<br />$
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.

asterisk_man · Nov 9, 2007, 11:50 AM

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 :)

asterisk_man · Nov 9, 2007, 12:06 PM

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.

helpless06 · Nov 9, 2007, 12:23 PM

This is the whole problem.

galactus · Nov 10, 2007, 06:28 AM

I would use tech to solve this monster.

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