# 81^x-2

• Oct 19th 2007, 09:52 AM
Danielisew
81^x-2
Question is in topic.
You must express this in form 9^y
I managed to do 1/9^1-x, and got 9^x as my answer (not sure if that's right), but cannot do this question :confused:

EDIT: Also, how would you do 3^4x+6, still expressing the answer in the form 9^y?

Thanks!
• Oct 19th 2007, 10:16 AM
topsquark
Quote:

Originally Posted by Danielisew
Question is in topic.
You must express this in form 9^y
I managed to do 1/9^1-x, and got 9^x as my answer (not sure if that's right), but cannot do this question :confused:

EDIT: Also, how would you do 3^4x+6, still expressing the answer in the form 9^y?

Thanks!

"Question is in topic." What topic?

Also: you need to use parenthesis. (Shake)

For the first problem:
$\frac{1}{9^{1 - x}} = 9^{-(1 - x)} = 9^{x - 1}$

For the second problem:
$3^{4x+6}$

Note that $9 = 3^{1/2}$, so
$3^{4x+6} = (9^{1/2})^{4x + 6} = 9^{\frac{4x + 6}{2}} = 9^{2x + 3}$

and what does any of this have to do with $81 - x^2$???

-Dan
• Oct 19th 2007, 10:17 AM
CaptainBlack
Quote:

Originally Posted by Danielisew
Question is in topic.
You must express this in form 9^y
I managed to do 1/9^1-x, and got 9^x as my answer (not sure if that's right), but cannot do this question :confused:

EDIT: Also, how would you do 3^4x+6, still expressing the answer in the form 9^y?

Thanks!

Try putting in the missing brackets to make your meaning clear.

RonL
• Oct 19th 2007, 11:11 AM
SnipedYou
For the first one notice that $81=9^{2}$

So we get $9^{2(x-2)}$

Distribute and get $9^{2x-4}$