# Solving for x in an exponential equation

• Feb 14th 2013, 05:51 PM
takaj
Solving for x in an exponential equation
Hello, one question has proved ​ insurmountable to me. Here it is:
(3/4)^x =(27/64)^(2x-3)

Here is what I have tried:

(3)^x/(4)^x=(27)^(2x-3)/(64)^(2x-3)

Then i cross multiplied to get...

(192)^(2x^2-3x)=(108)^(2x^2-3x)

I know that I need to make the bases the same, but i can't see how to do that at this point. Some guidance would be greatly appreciated.
• Feb 14th 2013, 06:02 PM
Plato
Re: Solving for x in an exponential equation
Quote:

Originally Posted by takaj
Hello, one question has proved ​ insurmountable to me. Here it is:
(3/4)^x =(27/64)^(2x-3)

$\displaystyle \left(\frac{27}{64}\right)^{2x-3}=\left(\frac{3}{4}\right)^{6x-9}$
• Feb 14th 2013, 06:56 PM
takaj
Re: Solving for x in an exponential equation
Quote:

Originally Posted by Plato
$\displaystyle \left(\frac{27}{64}\right)^{2x-3}=\left(\frac{3}{4}\right)^{6x-9}$

?
• Feb 14th 2013, 07:52 PM
Plato
Re: Solving for x in an exponential equation
Quote:

Originally Posted by takaj
?

The point is if $\displaystyle \left(\frac{3}{4}\right)^a=\left(\frac{3}{4}\right )^b$ then $\displaystyle a=b$.

I showed how the bases are the same.
• Feb 14th 2013, 08:13 PM
takaj
Re: Solving for x in an exponential equation
Thank you. I got it now.