# A simple algebra exercise

• Sep 17th 2010, 10:24 AM
klik11
A 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 AM
Wilmer
Apply these 2 rules:

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

a^(-p) = 1 / a^p
• Sep 17th 2010, 10:54 AM
undefined
Quote:

Originally Posted by klik11
$\displaystyle 3^{x} = 2$

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

The result is

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

How do I get to the answer?
Thanks!

Slightly different way that avoids logs.

$\displaystyle 27^{-x}=(3^3)^{-x}=3^{-3x}=(3^x)^{-3}=2^{-3}=\frac{1}{8}$
• Sep 17th 2010, 11:19 AM
klik11
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 AM
undefined
Quote:

Originally Posted by klik11
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

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

$\displaystyle 3^{2x+8}=3^{2x}\cdot3^8=9^x\cdot3^8$

continue..

Actually I get $\displaystyle (\frac{3}{b})^{8}$
• Sep 17th 2010, 11:30 AM
klik11
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 AM
undefined
Quote:

Originally Posted by klik11
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...

e^2x?

Continuing from where I left off above

$\displaystyle 9^x\cdot3^8=b^8$

$\displaystyle 9^x=(\frac{b}{3})^8$

$\displaystyle 9^{-x}=(\frac{3}{b})^8$
• Sep 17th 2010, 11:41 AM
klik11
3^2x *

Thanks!!!