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.

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)

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

Re: Solving for x in an exponential equation

Quote:

Originally Posted by

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

?

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.

Re: Solving for x in an exponential equation