please include your work? i just dont know how to:
1. Make the bases on each side of the equation the same.
2. Set the exponents equal to each other.
thank you very much!!!
$\displaystyle \displaystyle \sqrt {3}^{2x+4} = 9 ^{x-2} $
note that $\displaystyle 9 = 3^2 $ Moderator edit: and $\displaystyle \sqrt{3} = 3^{1/2}$
so you have
$\displaystyle \displaystyle {3^{\frac {2x+4}{2}}} = 3 ^{2x-4} $
so it's at the same base and now you can write that :
$\displaystyle \frac {2x+4}{2}} = 2x-4 $
can you continue ?
I think the previous poster misread the question.
$\displaystyle \left( {\sqrt 3 } \right)^{2x + 4} = \left( {3^{\frac{1}
{2}} } \right)^{2x + 4} = 3^{x + 2} $
$\displaystyle 9^{x - 2} = \left( {3^2 } \right)^{x - 2} = 3^{2x - 4} $
So $\displaystyle 3^{x + 2} = 3^{2x - 4} $
Can you finish?