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Math Help - Logs

  1. #1
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    Logs

    Solve for x: (9/4)log4x=10

    this is what I got:
    log4 (x^(9/4))=10
    4^10 = x^(9/4)
    (4^10)^(4/9) = (x^(9/4))^(4/9)
    4^(4/10) = x
    4^(2/5) =x

    am I on the right track?
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  2. #2
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    Re: Logs

    Quote Originally Posted by RD33 View Post
    Solve for x: (9/4)log4x=10

    this is what I got:
    log4 (x^(9/4))=10
    4^10 = x^(9/4)
    (4^10)^(4/9) = (x^(9/4))^(4/9)
    4^(4/10) = x
    4^(2/5) =x

    am I on the right track?
    First of all I presume this is log base 4? I'd recommend that you start this way:
    \frac{9}{4}log_4x = 10

    log_4x = \frac{4}{9} \cdot 10

    log_4x = \frac{40}{9}

    Can you work out the rest?

    -Dan
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  3. #3
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    Re: Logs

    I think so. I would have to put it into exponential from right? So I would end up with 4^(40/9) = x ?
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  4. #4
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    Re: Logs

    Hello, RD33!

    \text{Solve for }x\!:\;\;\tfrac{9}{4}\log_4x\:=\:10

    This is what I got:

    . . \log_4\left(x^{\frac{9}{4}}\right) \:=\:10

    . . 4^{10} \:=\: x^{\frac{9}{4}}

    . . \left(4^{10}\right)^{\frac{4}{9}} \:=\: \left(x^{\frac{9}{4}}\right)^{\frac{4}{9}}

    . . 4^{\color{red}{\frac{4}{10}}} \:=\: x \;\;\color{red}{??}

    . . 4^{\frac{2}{5}}\:=\: x

    Am I on the right track?
    Your reasoning is excellent.
    Just one error in arithmetic.

    I solved it like this:

    . . \tfrac{9}{4}\log_4x \:=\:10

    . . . \log_4x \:=\:\tfrac{40}{9}

    . . . . . . x \:=\:4^{\frac{40}{9}}
    Thanks from RD33
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  5. #5
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    Re: Logs

    Ok thank you
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