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Math Help - Solving an exponential equation - Check my method

  1. #1
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    Solving an exponential equation - Check my method

    Hi,

    I'm new to the forum, but decided to give it a try now, because I can't seem to get around what I'm doing wrong. I'm revising for my IB finals and can't seem to get this one question down, which is from a past final exam:

    So the question is: Solve the equation 4^(x-1)=2^x+8

    My method:
    4^(x-1) = 2^x+8
    4^(x-1) = 2^x+2^3
    2^(2(x-1)) = 2^x+2^3
    2^(2x-2) = 2^x+2^3 | log
    log 2 (2^(2x-2)) = log 2 (2^x) + log 2 (8)
    2x-2(log 2 (2)) = x (log 2 (2)) + 3
    2x-2 = x + 3
    2x-x = 3 + 2
    x = 5

    But this doesn't check out with the original equation, because it comes to 256 = 40
    It's been a while since we did logarithms and exponents, so I might be doing something incorrect. So please correct me where I'm wrong.

    Thanks in advance.
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  2. #2
    Super Member Quacky's Avatar
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    Quote Originally Posted by FinnSkies View Post
    Hi,

    I'm new to the forum, but decided to give it a try now, because I can't seem to get around what I'm doing wrong. I'm revising for my IB finals and can't seem to get this one question down, which is from a past final exam:

    So the question is: Solve the equation 4^{x-1}=2^x+8

    My method:
    4^(x-1) = 2^x+8
    4^(x-1) = 2^x+2^3
    2^(2(x-1)) = 2^x+2^3
    2^(2x-2) = 2^x+2^3
    log 2 (2^(2x-2)) = log 2 (2^x) + log 2 (8) Unfortunately, no. This is the mistake. Log(a+b)\neq Log(a)+Log(b) Taking logs of both sides will lead you to a dead end here. Review this stage carefully, you've not taken logs of the right hand side correctly.
    Starting at this stage: 2^{2x-2} = 2^x+2^3

    Write it like this:

    \displaystyle\frac{(2^{2x})}{2^2}-2^x-2^3=0

    Let 2^x=Y

    \displaystyle\frac{Y^2}{4}-Y-8=0

    And solve for Y
    Last edited by Quacky; March 10th 2011 at 02:18 PM.
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  3. #3
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    Hello, FinnSkies!

    Another approach . . .


    \text{Solve: }\:4^{x-1} \:=\:2^x+8

    \text{My method:}
    . . . . 4^{x-1} \:=\: 2^x+8
    . . (2^2)^{x-1} \:=\: 2^x+2^3
    . . . 2^{2(x-1)} \:=\: 2^x+2^3
    . . . . 2^{2x-2} \:=\: 2^x+2^3 . . Correct!

    We have: . 2^{2x-2} - 2^x - 8 \:=\:0

    I saw that I could turn it into a quadratic if I multiplied by 2^2\!:

    . . 2^2\!\cdot\!2^{2x-2} - 2^2\!\cdot\!2^x - 2^2\!\cdot\!8 \:=\:0 \quad\Rightarrow\quad 2^{2x} - 4\!\cdot\!2^x - 32 \:=\:0


    Factor: . (2^x + 4)(2^x - 8) \:=\:0


    Then we have: . 2^x+4\:=\:0 \quad\Rightarrow\quad 2^x \:=\:-4\quad\text{no real roots}

    . . . . . . . . . . . 2^x - 8 \:=\:0 \quad\Rightarrow\quad 2^x \:=\:8 \quad\Rightarrow\quad \boxed{x \:=\:3}

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