Hi,

I cannot solve the following equation to x.

$\displaystyle 27^{\sqrt{2-3x}} = 243^{\sqrt{3x-2}}$

I tried it like this:

$\displaystyle \sqrt{2-3x}lg(27) = \sqrt{3x-2}lg(243)$

after some squaring and multiplying I got the following:

$\displaystyle 2-3x = \frac{3x(lg(243))^2} {(lg(27))^2} - \frac{2(lg(243))^2)}{(lg(27))^2}$

Then I didn't know how to continue. Wolfram Alpha says that x is 2/3:

http://www.wolframalpha.com/input/?i=2-3x+%3D+%28lg%28243%29%2F%28lg%2827%29%29%29^2*%283 x-2%29
My textbook, from where I have this equation, also says x = 2/3. But I don't know how to get there. I hope that someone can post the last steps to solve to x.