$\displaystyle 3^{x} = 2$

$\displaystyle 27^{-x} = ?$

The result is

$\displaystyle \dfrac{1}{8}$

How do I get to the answer?

Thanks!

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- Sep 17th 2010, 10:24 AMklik11A simple algebra exercise
$\displaystyle 3^{x} = 2$

$\displaystyle 27^{-x} = ?$

The result is

$\displaystyle \dfrac{1}{8}$

How do I get to the answer?

Thanks! - Sep 17th 2010, 10:51 AMWilmer
Apply these 2 rules:

If a^p = b then p = log(b) / log(a)

a^(-p) = 1 / a^p - Sep 17th 2010, 10:54 AMundefined
- Sep 17th 2010, 11:19 AMklik11
Thanks!

I have one more exercise like this though, and I still can't solve it.

$\displaystyle 3^{2x+8} = b^{8}$

$\displaystyle 9^{-x} = ?$

the result is

$\displaystyle (\frac{b}{3})^{8}$ - Sep 17th 2010, 11:24 AMundefined
- Sep 17th 2010, 11:30 AMklik11
Yes my mistake, it's

$\displaystyle (\frac{3}{b})^{8}$

Actually, I already figured that 9^x equals e^2x. I have no idea how to continue the exercise... - Sep 17th 2010, 11:37 AMundefined
- Sep 17th 2010, 11:41 AMklik11
3^2x *

Thanks!!!