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  1. Oct 2, 2013 #1
    1. The problem statement, all variables and given/known data
    4^x + 6(4^-x) = 5

    2. Relevant equations
    log??? since this is the unit we are doing, but i'm not sure if it applies.

    3. The attempt at a solution

    I was thinking of changing 4^x to a variable like 'a', but 4^-x is not the same so I can't replace that with 'a' as well.
    Then I though 4^-x = 1/4 ^x
    4^x + 6[(1/4)^x] = 5

    and I'm lost.
    thanks :)
  2. jcsd
  3. Oct 2, 2013 #2

    Simon Bridge

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    You have:

    $$4^{x}+6(4^{-x})=5$$ ... is that right?

    ... and you want to put it into some form that you can integrate or differentiate or something?
    You suspect it has something to do with taking a logarithm - so do I :), since ##\log_4(4^x)=x## hugely simplifies the problem.

    Your intuition to put ##a=4^x## is a good one - with that substitution you get:
    $$a+\frac{6}{a}=5$$ ... which is hard to think about, so put it in standard form.
    Hint: multiply both sides by ##a##.
    What sort of equation is that?
    Last edited: Oct 2, 2013
  4. Oct 2, 2013 #3
    it is actually,

    4^x + 6(1/4)^x = 5
  5. Oct 2, 2013 #4
    Quadratic :) thankyou
  6. Oct 3, 2013 #5

    Simon Bridge

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    Well done :)
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