1. ## solve for x??

1. Make the bases on each side of the equation the same.

2. Set the exponents equal to each other.

thank you very much!!!

2. Originally Posted by lollilikeslife

1. Make the bases on each side of the equation the same.

2. Set the exponents equal to each other.

thank you very much!!!

$\displaystyle \sqrt {3}^{2x+4} = 9 ^{x-2}$

note that $9 = 3^2$ Moderator edit: and $\sqrt{3} = 3^{1/2}$

so you have

$\displaystyle {3^{\frac {2x+4}{2}}} = 3 ^{2x-4}$

so it's at the same base and now you can write that :

$\frac {2x+4}{2}} = 2x-4$

can you continue ?

3. I think the previous poster misread the question.
$\left( {\sqrt 3 } \right)^{2x + 4} = \left( {3^{\frac{1}
{2}} } \right)^{2x + 4} = 3^{x + 2}$

$9^{x - 2} = \left( {3^2 } \right)^{x - 2} = 3^{2x - 4}$

So $3^{x + 2} = 3^{2x - 4}$

Can you finish?

4. Originally Posted by Plato
I think the previous poster misread the question.
$\left( {\sqrt 3 } \right)^{2x + 4} = \left( {3^{\frac{1}
{2}} } \right)^{2x + 4} = 3^{x + 2}$

$9^{x - 2} = \left( {3^2 } \right)^{x - 2} = 3^{2x - 4}$

So $3^{x + 2} = 3^{2x - 4}$

Can you finish?
we wrote the same thing i didn't misread anything

5. Originally Posted by yeKciM
we wrote the same thing i didn't misread anything
The expression you posted:
Originally Posted by yeKciM
$\displaystyle \sqrt {3^{2x+4}} = 9 ^{x-2}$
[snip]
suggested otherwise. I assume by accident you were square rooting everything rather than just the 3. I have edited your original post.