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Math Help - logarithm equations! need a little help solving...

  1. #1
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    logarithm equations! need a little help solving...

    i'm having trouble solving a few of these logarithm problems

    1) If loga2 = logb16, show that b=a^4

    i'm also having trouble with squaring logs...for example one of the questions was:
    (log10x)^2 - 2log10x + 1 = 0
    the answer is 10, but i keep ending up canceling out both xs

    the last one was(2log2x)^2 - 3log2x - 10 = 0
    how exactly do you end up with two values for x in that example? the answers are x = 4, 2^(-5/4)

    thanks!!
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  2. #2
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    \displaystyle \log_y{x}=\frac{\log{x}}{\log{y}}

    \displaystyle \frac{\log{2}}{\log{a}}=\frac{\log{16}}{\log{b}}\R  ightarrow \frac{\log{2}}{4\log{2}}=\frac{\log{a}}{\log{b}}

    \displaystyle \frac{1}{4}=\frac{\log{a}}{\log{b}}\Rightarrow \frac{1}{4}\log{b}=\log{a}\Rightarrow 10^{\log{b^{1/4}}}10^{\log{a}}

    \displaystyle b^{1/4}=a\Rightarrow b=a^4
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  3. #3
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    Do you mean this

    \displaystyle (\log_{10}{x})^2 - 2\log_{10}{x} + 1 = 0

    or this

    \displaystyle (\log{10x})^2 - 2\log{10x} + 1 = 0
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  4. #4
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    the first one, 10 is the base
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  5. #5
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    \displaystyle (\log_{10}{x})^2 - 2\log_{10}{x} + 1 = 0\Rightarrow (\log{x}-1)^2=0\Rightarrow \log{x}-1=0

    \mbox{Let} \ (\log{x})^2=y^2 \ \ \log{x}=y\Rightarrow y^2-2y+1=0\Rightarrow (x-1)^2=0 \ \mbox{but} \ y=\log{x}

    \Rightarrow\log{x}=1\Rightarrow 10^{log{x}}=10^1\Rightarrow x=10
    Last edited by dwsmith; December 24th 2010 at 08:30 PM. Reason: Adding line 2 for clarity and made a variable change
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  6. #6
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    (2\log_{2}{x})^2 - 3\log_{2}{x} - 10 = 0

    Is two the base?
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  7. #7
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    yes, 2 is the base
    thanks for the other answers ^_^
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  8. #8
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    (2\log_{2}{x})^2 - 3\log_{2}{x} - 10 = 0

    \mbox{Let} \ y=\log_{2}{x} \ \Rightarrow 4y^2-3y-10=0\Rightarrow (y-2)(4y+5)=0

    y=2 \ \mbox{and} \ \frac{-5}{4}

    \displaystyle \log_2{x}=2\Rightarrow 2^2=x\Rightarrow x=4

    \displaystyle \log_2{x}=\frac{-5}{4}\Rightarrow 2^{-5/4}=x\Rightarrow x=2^{-5/4}
    Last edited by dwsmith; December 24th 2010 at 08:45 PM.
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  9. #9
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    thanks!!
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  10. #10
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    Your welcome. Whenever you have logs of the same base in a quadratic appearance, try to factor it.
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