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
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: . . $\displaystyle 8-4n \;=\;-2n - 2$
. . . . . . . . . . . $\displaystyle -2n \;=\;-10$
. . . . . . . . . . . . . $\displaystyle n \;=\;5$
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} $