1. ## 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

2. Hello, GoodLuckMe!

The procedure is:

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

I'll do the last one.

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

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

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

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

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

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

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

. . . . . . . . . . . . . $\displaystyle n \;=\;5$

3. I'll do the first one (another example)

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

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

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

Make the bases the same:

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

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

Now cancel

$\displaystyle -2n = 3$

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

4. thanks-that's wonderfully helpful now I've got the method It should be alot easier.

much appreciated.