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Math Help - Logarithms!

  1. #1
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    Logarithms!

    I need some help solving the following:

    a) (2)*3^x = (7)*5^x

    b) 12^x = (4)*8^(2x)

    c)(12)*6^(2x-1) = 11^(x+3)

    d) (7)*0.43^(2x) = (9)*6^(-x)

    Thanks.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by andymac View Post
    I need some help solving the following:

    a) (2)*3^x = (7)*5^x

    b) 12^x = (4)*8^(2x)

    c)(12)*6^(2x-1) = 11^(x+3)

    d) (7)*0.43^(2x) = (9)*6^(-x)

    Thanks.
    Two methods for these. I will show you both for the first problem and you can apply it to the rest

    Method one

    2\cdot{3^x}=7\cdot{5^{x}}

    taking the natural log of both sides gives

    ln(2\cdot{3^{x}})=\ln(2)+\ln(3)x=\ln(7\cdot{5^x})=  \ln(7)+\ln(5)x

    Algebraically manipulating gives

    ln(3)x-\ln(5)x=x(\ln(3)-\ln(5))=x\ln\bigg(\frac{3}{5}\bigg)=\ln(7)-\ln(2)=\ln\bigg(\frac{7}{2}\bigg)

    so x=\frac{\ln\bigg(\frac{7}{2}\bigg)}{\ln\bigg(\frac  {3}{5}\bigg)}

    Method two

    2\cdot{3^x}=7\cdot{5^x}

    through manipulation we get

    \frac{3^x}{5^x}=\bigg(\frac{3}{5}\bigg)^x=\frac{7}  {2}

    taking the natural log of both sides gives

    x\ln\bigg(\frac{3}{5}\bigg)=\ln\bigg(\frac{7}{2}\b  igg)\Rightarrow{x=\frac{\ln\bigg(\frac{7}{2}\bigg)  }{\ln\bigg(\frac{3}{5}\bigg)}}
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