Indices help needed

• Sep 4th 2010, 02:54 AM
GoodLuckMe
Indices help needed
Just some hard indices problems I need help with, I have no idea how to go about these:

find n:

(1/4)^n=8

(125/8)^n=25/4

(81/16)^n=32/243

81^2n-1=27^3n+1

16^2-n = (1/4)n+1

thanks, as I can't seem to find the right method
• Sep 4th 2010, 04:28 AM
Soroban
Hello, GoodLuckMe!

The procedure is:

. . Express all exponential in the same base.
. . Equate exponents.
. . Solve for $n.$

I'll do the last one.

Quote:

$16^{2-n} \:=\: \left(\frac{1}{4}\right)^{n+1}$

We note that: . $16 = 2^4\,\text{ and }\,4=2^2$

So we have: . $\left(2^4\right)^{2-n} \;=\;\left(\frac{1}{2^2}\right)^{n+1}$

. . . . . . . . . . $2^{4(2-n)} \;=\;\left(2^{-2}\right)^{n+1}$

. . . . . . . . . . . $2^{8=4n} \;=\;2^{-2n-2}$

. . . Then: . . $8-4n \;=\;-2n - 2$

. . . . . . . . . . . $-2n \;=\;-10$

. . . . . . . . . . . . . $n \;=\;5$
• Sep 4th 2010, 06:54 AM
jgv115
I'll do the first one (another example)

$(\frac{1}{4})^n=8$

Note that $(\frac{1}{4})^n = 4^{-n}$

So... $4^{-n} = 8$

Make the bases the same:

$2^{2*-n} = 2^{3*1}$

$2^{-2n} = 2^3$

Now cancel

$-2n = 3$

$n= \frac{-3}{2}$

:)
• Sep 5th 2010, 01:07 PM
GoodLuckMe
thanks-that's wonderfully helpful :) now I've got the method It should be alot easier.

much appreciated.