# Thread: Help with simple equation needed.

1. ## Help with simple equation needed.

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.

2. Originally Posted by Antamara
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.
Why didn't you try

$\displaystyle 3^{3\sqrt{2-3x}}=3^{5\sqrt{3x-2}}$ ?

Then equate the indices and square both of them

3. Do you "see" (from Archie's reply) that 3SQRT(2 - 3x) = 5SQRT(3x - 2) ?

4. Thank you very much for the fast answers. I tried it with Archie Meades' recommendation and it works. I also tried to continue from here:

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

I transformed the above equation to this:

$\displaystyle x = (\frac{-2(lg243)^2}{(lg27)^2}-2) : (\frac{-3(lg(243))^2} {(lg27)^2}-3)$

Which gives x = 2/3.

5. Notice that, as Archie Meade pointed out, $\displaystyle 27= 3^3$ while $\displaystyle 243= 3^5$. That means that $\displaystyle \frac{log(243)}{log(27)}= \frac{log(3^5)}{log(3^3)}$$\displaystyle = \frac{5 log(3)}{3 log(3)}= \frac{5}{3}$.