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Math Help - Using an Exponent to Undo Log

  1. #1
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    Using an Exponent to Undo Log

    What is the next step in solving this (using this method)? But there is another way which involves using the log laws (which I know how to do).

    \log x - \dfrac{1}{3}\log8 = \log7

    10^{\log(x)- \frac{1}{3}\log8} = 10^{\log7}

    I believe the next step involves dividing the bases of the exponents, since the exponents are subtracted.
    Last edited by Jason76; June 28th 2013 at 08:25 PM.
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  2. #2
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    Re: Using an Exponent to Undo Log

    why r u doing this???

    isnt log x=log7 +log2
    use log a + log b= log ab
    and find the answer
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  3. #3
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    Re: Using an Exponent to Undo Log

    An example of exponanted stuff:

    e^{\ln (2 + y^{2)}} = e^{\ln(4 + x^{2}) + C}

    2 + y^{2} = (4 + x^{2}) e^{C} Since adding, then multiplying comes in.

    But with the other example on this page, some subtracting was in there.
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  4. #4
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    Re: Using an Exponent to Undo Log

    Do it with the laws of logarithms, you will get the answer straight unless you have a special purpose.
    lox x - 1/3 log 8 = log 7
    log x - log(8)^(1/3 ) = log 7
    log x - log 2 = log 7
    log ( x/2) = log 7 that will give x = 14
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  5. #5
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    Re: Using an Exponent to Undo Log

    Hello, Jason76!

    \text{Solve for }x\!:\;\log x - \tfrac{1}{3}\log8 \:=\: \log7

    Simplify the logs before you exponentiate.

    We have: . \log x \:=\:\log 7 + \tfrac{1}{3}\log 8

    . . . . . . . . \log x \;=\;\log 7 + \log\left(8^{\frac{1}{3}}\right)

    . . . . . . . . \log x \;=\;\log 7 + \log 2

    . . . . . . . . \log x \;=\;\log(7\cdot2)

    . . . . . . . . \log x \;=\;\log 14

    Therefore: n . x \;=\;14
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