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Math Help - A log problem

  1. #1
    Junior Member
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    A log problem

    Can someone please show me step by step how to solve down on this log problem...

    8^-2x+4 = 64
    log9x = 1/2
    log3 (x-1) + log3 (x+1) = 1

    Thank you very much.
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  2. #2
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    Re: A log problem

    I assume the first problem is really 8^(-2x+4)=64

    take log8 of both sides

    log8(8-2x+4)=log8(64)

    which leads to

    (-2x+4)log8(8)=2log8(8)

    (-2x+4) = 2

    x = 1


    second problem

    I would first rewrite log9x=1/2 as (ln(x))/(ln(9))=1/2

    then

    ln(x) = (1/2) ln(9)

    ln(x) = ln(91/2)

    x = 91/2=3


    third

    first rewrite as

    (ln(x-1) + ln(x+1))/(ln3)=1

    ln(x-1)+ln(x+1) = ln3
    ln(x2-1)=ln3
    x2-1=3
    x2=4
    x=2

    x can't be -2 because it would cause you to take the log of a negative number

    ln=natural log
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  3. #3
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    Re: A log problem

    Quote Originally Posted by bmoon123 View Post
    second problem

    I would first rewrite log9x=1/2 as (ln(x))/(ln(9))=1/2

    then

    ln(x) = (1/2) ln(9)

    ln(x) = ln(91/2)

    x = 91/2=3


    third

    first rewrite as

    (ln(x-1) + ln(x+1))/(ln3)=1

    ln(x-1)+ln(x+1) = ln3
    ln(x2-1)=ln3
    x2-1=3
    x2=4
    x=2

    x can't be -2 because it would cause you to take the log of a negative number

    ln=natural log
    It's easier to just go

    \displaystyle \begin{align*} \log_9{x} &= \frac{1}{2} \\ x &= 9^{\frac{1}{2}} \\ x &= 3 \end{align*}

    and

    \displaystyle \begin{align*} \log_3{(x - 1)} + \log_3{(x + 1)} &= 1 \\ \log_3{(x - 1)(x + 1)} &= 1 \\ (x - 1)(x + 1) &= 3^1 \\ x^2 - 1 &= 3 \\ x^2 &= 4 \\ x &= \pm 2 \\ x &= 2 \textrm{ due to the fact you can only take the logarithm of a positive number} \end{align*}
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