1. ## [SOLVED] Eponential Equations

I need to solve for x in the equation:
9^(2x+1)=81(27^2)
I've tried various things, getting every answer but the correct one, which is 2. I have no idea how x=2 though.
Here's one of my (futile) tries:
9^(2x+1)=9^2(27^x)
x+1=27^x
1 = (27^xx)
Guidance through the proper steps would be appreciated, thank you!

2. Originally Posted by ria3
I need to solve for x in the equation:
9^(2x+1)=81(27^2)
I've tried various things, getting every answer but the correct one, which is 2. I have no idea how x=2 though.
Here's one of my (futile) tries:
9^(2x+1)=9^2(27^x)
x+1=27^x
1 = (27^xx)
Guidance through the proper steps would be appreciated, thank you!
So:
$\displaystyle 9^(2x+1)=81$x$\displaystyle 27^2$

Put everything as powers of 3:
$\displaystyle (3^2)^(2x+1)=(3^4)(3^3)^2$
=$\displaystyle 3^(4x+2)=(3^4)(3^6)$
=$\displaystyle 3^(4x+2)=3^10$
therefore: $\displaystyle 4x+2=10, 4x=8, x=2$

Sorry for the strange powers, I'm still trying to get the hang of this math code stuff. =)

3. thank you!