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Math Help - Solve the equation

  1. #1
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    Solve the equation

    Solve the equation
    4^(2x)-4^(x-1)-12=0

    How is this equation solved using logs?? Please help.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by haku View Post
    Solve the equation
    4^(2x)-4^(x-1)-12=0

    How is this equation solved using logs?? Please help.
    Rewrite this as:

    4^{2x}-4^{x-1}-12 = (4^x)^2 - (4^x)/4 -12=0

    Then change the variable to y=4^x, so that x=\log_4(y). Then:

    y^2 - y/4-12=0

    Now solve for y, then for x

    RonL
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  3. #3
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    4^(2x) -4^(x-1) -12 = 0

    Cannot take the logs of both sides because log(0) is undefined or there are no log(0). So,

    4^(2x) -4^(x-1) = 12

    Take the logs of both sides,
    2x*log(4) -(x-1)log(4) = log(12)
    2x*log(4) -x*log(4) +log(4) = log(4*3)
    x*log(4) +log(4) = log(4) +log(3)
    x*log(4) = log(3)
    x = log(3) / log(4) --------------answer.

    Or, x = (1/2)[log(3) / log(2)]
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  4. #4
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    Thanks for the help guys!
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  5. #5
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    Hello, haku!

    Solve: . 4^{2x} - 4^{x-1} - 12\:=\:0

    We have: . 4^{2x} - 4^x\!\cdot 4^{-1} - 12 \;=\;0 \quad\Rightarrow\quad 4^{2x} - \frac{1}{4}\!\cdot\!4^x - 12 \;=\;0

    Multiply by 4: . 4\cdot4^{2x} - 4^x - 48 \;=\;0

    We have a quadratic: . 4(4^x)^2 - 4^x - 48 \;=\;0

    Quadratic Formula: . 4^x\;=\;\frac{1 +\sqrt{769}}{8}

    Then: . x \:=\:\log_4\left(\frac{1 + \sqrt{769}}{8}\right) \;=\;\frac{\ln\left(\frac{1+\sqrt{769}}{8}\right)}  {\ln(4)} \;\approx\;0.922264366

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  6. #6
    Math Engineering Student
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    Quote Originally Posted by ticbol View Post
    4^(2x) -4^(x-1) = 12

    Take the logs of both sides,
    2x*log(4) -(x-1)log(4) = log(12)
    You made a mistake here
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  7. #7
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    Quote Originally Posted by Krizalid View Post
    You made a mistake here
    Yes.

    [I just took a peek. I'm home for a while.]

    Log of the original LHS is not as I wrte it. I fell on the trap.
    log(a +b) is not log(a) +log(b), as I warned in another reply before.
    And, normally I check my answer against the original equation, quickly or in "scratch paper" or calculator, before I say that is my answer. I forgot to do that this morning before I left for work.
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