# Solving this equation

• Oct 16th 2011, 06:47 AM
Burbujas
Solving this equation
Can anyone help me solve this basic equation (including the steps). I know the answer, I just don't know how to get there.

5^9x+2 = (1/25)^2x-9

Excuse the format, don't know how to do exponents on this forum.

Thanks a lot.
• Oct 16th 2011, 06:52 AM
e^(i*pi)
Re: Solving this equation
Quote:

Originally Posted by Burbujas
Can anyone help me solve this basic equation (including the steps). I know the answer, I just don't know how to get there.

5^9x+2 = (1/25)^2x-9

Excuse the format, don't know how to do exponents on this forum.

Thanks a lot.

I assume the question is $\displaystyle 5^{9x+2} = \left(\dfrac{1}{25}\right)^{2x-9}$

Note that $\displaystyle \dfrac{1}{25} = 5^{-2}$ so the RHS becomes $\displaystyle 5^{-2(2x-9)}$

If two bases are the same then their exponents must be the same: $\displaystyle 9x+2 = -2(2x-9)$. Solve for x
• Oct 16th 2011, 08:03 AM
Burbujas
Re: Solving this equation
Oh wow, that was so simple. I got it now, thanks a lot.