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Thread: Indices help needed

  1. #1
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    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
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  2. #2
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    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: . . $\displaystyle 8-4n \;=\;-2n - 2$

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

    . . . . . . . . . . . . . $\displaystyle n \;=\;5$
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  3. #3
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    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} $

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

    much appreciated.
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