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Thread: logarithims

  1. #1
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    logarithims

    i'm getting stuck ....a little help please

    $\displaystyle 5^{x-2} = 4^x$

    i got..
    $\displaystyle log 5 ^{x-2} = log 4^x$
    $\displaystyle (x-2) log 5 = x log 4 $

    and i'm stuck here...
    Last edited by extraordinarymachine; Dec 17th 2009 at 02:33 AM.
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  2. #2
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    Quote Originally Posted by extraordinarymachine View Post
    i'm getting stuck ....a little help please

    $\displaystyle 5^x-2 = 4^x$

    i got..
    $\displaystyle log 5 ^x-2 = log 4^x$ <<<<<<<<< never commit such a crime again!
    $\displaystyle (x-2) log 5 = x log 4 $

    and i'm stuck here...
    Do you mean

    $\displaystyle 5^x - 2 = 4^x$ If so

    $\displaystyle \left(\dfrac54 \right)^x = 2~\implies~x=\log_{\frac54}(2)=\dfrac{\ln(2)}{\ln( 5)-\ln(4)}$

    or

    $\displaystyle 5^{x-2} = 4^x$ If so

    $\displaystyle \frac1{25} \cdot 5^x = 4^x~\implies~\left(\dfrac54 \right)^x = 25 ~\implies~x=\log_{\frac54}(25)=\dfrac{\ln(25)}{\ln (5)-\ln(4)}$
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  3. #3
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    Quote Originally Posted by earboth View Post
    Do you mean

    $\displaystyle 5^x - 2 = 4^x$ If so

    sorry i messed up the question, i ment to write it like this

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  4. #4
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    Smile

    Quote Originally Posted by extraordinarymachine View Post
    i'm getting stuck ....a little help please

    $\displaystyle 5^{x-2} = 4^x$

    i got..
    $\displaystyle log 5 ^{x-2} = log 4^x$
    $\displaystyle (x-2) log 5 = x log 4 $

    and i'm stuck here...
    $\displaystyle (x-2) log 5 = x log 4
    \Leftrightarrow \frac{ x-2}{x }$$\displaystyle = \frac{log4 }{log5 } \Leftrightarrow 1- \frac{ 2}{x } =\frac{log4 }{log5 } $
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  5. #5
    Senior Member Stroodle's Avatar
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    $\displaystyle (x-2)log5=xlog4$

    $\displaystyle \frac{x-2}{x}=\frac{log4}{log2}$

    $\displaystyle 1-\frac{2}{x}=\frac{log4}{log5}$

    $\displaystyle \frac{2}{x}=1-\frac{log4}{log5}$

    $\displaystyle \frac{2}{x}=\frac{log5}{log5}-\frac{log4}{log5}$

    $\displaystyle \frac{2}{x}=\frac{log5-log4}{log5}$

    $\displaystyle \frac{x}{2}=\frac{log5}{log5-log4}$

    $\displaystyle x=\frac{2log5}{log5-log4}$
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  6. #6
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    5 steps
    $\displaystyle (x-2)log5=xlog4$
    $\displaystyle xlog5-2log5=xlog4$
    $\displaystyle xlog5-xlog4=2log5$
    $\displaystyle x(log5-log4)=2log5$
    $\displaystyle x=\frac{2log5}{log5-log4}$
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  7. #7
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    Quote Originally Posted by Krahl View Post
    5 steps
    $\displaystyle (x-2)log5=xlog4$
    $\displaystyle xlog5-2log5=xlog4$
    $\displaystyle xlog5-xlog4=2log5$
    $\displaystyle x(log5-log4)=2log5$
    $\displaystyle x=\frac{2log5}{log5-log4}$
    this was really helpful thanks!
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