# Solving for x in equations involving logs

• Jul 28th 2010, 04:09 PM
jkrowling
Solving for x in equations involving logs
Could somebody point me in the right direction?

$e^x + e^-^x = 2$
$e^x (e^x + e^-^x) = 2*e^x$
$e^2x + e = 2e^x$
$e^2x - 2e^x + e = 0$

Something tells me I should be able to solve this as a quadratic, but e is a constant, and I can't figure out how to do this with a constant as the base.
• Jul 28th 2010, 04:17 PM
pickslides
$e^x+e^{-x}=2$

Times both sides by $e^x$

$e^{2x}+1=2 e^x$

$e^{2x}-2 e^x+1=0$

make $e^x = a$ giving

$a^{2}-2 a+1=0$

Solve from here?
• Jul 28th 2010, 04:27 PM
jkrowling
Ok, but why is $e^x * e^-^x$ =1? Why isn't it $e$?
• Jul 28th 2010, 04:33 PM
pickslides
$e^x\times e^{-x}= e^{x-x}= e^0 = 1$