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Math Help - Incorrect Logarithm, please help me fix it.

  1. #1
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    Incorrect Logarithm, please help me fix it.

    4^(x + 2) = 200


    ln(4^(x+2) = ln(200)


    xln(4) + 2ln(4) = ln(200)


    xln(4) = -2ln(4) + ln(200)


    xln(4)/ln(4) = (-2ln(4) + ln(200))*1/ln(4)


    x = (-2ln(4) + ln(200))* 1/ln(4)


    x = -2ln(4) + ln(200)/ln(4)


    = 1.04

    What am I doing wrong?
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  2. #2
    Junior Member
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    Quote Originally Posted by bobbyboy1111 View Post

    x = (-2ln(4) + ln(200))* 1/ln(4)


    x = -2ln(4) + ln(200)/ln(4)

    What am I doing wrong?
    If you are going to divide the RHS by ln(4), then the -2ln(4) term becomes just -2.

    You removed the parenthesis but didn't cancel out that term.

    Btw, the easier way to solve this is just

    log_4 (200) = x + 2

    Then use change of base formula to compute log_4 (200). You'll get the same answer either way, but save a few steps.
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  3. #3
    Senior Member Twig's Avatar
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    Hi!

    4^{x+2}=200

    (x+2)\cdot ln(4)=ln(200)

    x\cdot ln(4)=ln(200)-2\cdot ln(4)

     \Rightarrow \; x = \frac{ln(200)-2\cdot ln(4)}{ln(4)}


    Note: This can be simplified etc, but I didnt bother there, hoping it is easier to follow.
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