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Math Help - Exponential Equation

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    Exponential Equation



    Where to go from here?
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    Re: Exponential Equation

    Quote Originally Posted by Remriel View Post


    Where to go from here?
    \displaystyle \begin{align*} \frac{x + 1}{x} &= \frac{\log{5}}{\log{4}} \\ 1 + \frac{1}{x} &= \frac{\log{5}}{\log{4}} \\ \frac{1}{x} &= \frac{\log{5}}{\log{4}} - 1 \\ \frac{1}{x} &= \frac{\log{5} - \log{4}}{\log{4}} \\ x &= \frac{\log{4}}{\log{5} - \log{4}} \end{align*}
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  3. #3
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    Re: Exponential Equation

    Quote Originally Posted by Remriel View Post


    Where to go from here?
    Your calculations are OK, but there is a simpler way to do this question:

    \displaystyle{5^x=4^{x+1}~\implies~5^x=4 \cdot 4^{x}~\implies~\left(\frac54 \right)^x=4~\implies~x = \log_{\frac54}(4)}

    With your result:

    \displaystyle{(x)log5=(x+1)log4~\implies~(x)log5=(  x)log4 + log4~\implies~(x)log5-(x)log4 = log4~\implies~x=\frac{log4}{log5-log4}}
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    Re: Exponential Equation

    Go back to (x)log5=(x+1)log4
    So (x)log5=(x)log4+log4
    x(log5-log4)=log4
    x=(log4)/(log5-log4}
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    Re: Exponential Equation

    What about this one? I can't find a way to separate:
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    Re: Exponential Equation

    Quote Originally Posted by Remriel View Post
    What about this one? I can't find a way to separate:
    \displaystyle \begin{align*} \frac{3x + 1}{x - 2} &= \frac{\log{3}}{\log{2}} \\ \frac{3(x - 2) + 7}{x - 2} &= \frac{\log{3}}{\log{2}} \\ 3 + \frac{7}{x - 2} &= \frac{\log{3}}{\log{2}} \end{align*}

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    Re: Exponential Equation

    Is this correct?


    Last edited by Remriel; July 2nd 2012 at 04:06 PM.
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    Re: Exponential Equation

    \frac{3x+1}{x-2} = \frac{\log{3}}{\log{2}}

    cross multiply ...

    (3x+1)\log{2} = (x-2)\log{3}

    distribute both sides ...

    3x\log{2} + \log{2} = x\log{3} - 2\log{3}

    get all x terms on the same side ...

    3x\log{2} - x\log{3} = -\log{2} - 2\log{3}

    factor out the x ...

    x(3\log{2} - \log{3}) = -(\log{2}+2\log{3})

    x\log\left(\frac{8}{3}\right) = -\log{18}

    solve for x ...

    x = -\frac{\log{18}}{\log\left(\frac{8}{3}\right)}
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    Re: Exponential Equation

    Quote Originally Posted by Remriel View Post
    Is this correct?


    No, when you have x in the denominator, you need to reciprocate. It should be like this...

    \displaystyle \begin{align*} 3 + \frac{7}{x - 2} &= \frac{\log{3}}{\log{2}} \\ \frac{7}{x - 2} &= \frac{\log{3}}{\log{2}} - 3 \\ \frac{7}{x - 2} &= \frac{\log{3} - 3\log{2}}{\log{2}} \\ \frac{x - 2}{7} &= \frac{\log{2}}{\log{3} - 3\log{2}} \\ x - 2 &= \frac{7\log{2}}{\log{3} - 3\log{2}} \\ x &= \frac{7\log{2}}{\log{3} - 3\log{2}} + 2 \\ x &= \frac{\log{2} + 2\log{3}}{\log{3} - 3\log{2}} \end{align*}
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