# Thread: 81^x-2

1. ## 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

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

Thanks!

2. 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

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.

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

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

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

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

-Dan

3. 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

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

4. For the first one notice that $\displaystyle 81=9^{2}$

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

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