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Math Help - Logarithmic Equation With Two Different Logs

  1. #1
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    Logarithmic Equation With Two Different Logs

    Solve the log equation below. Express irrational solutions in exact form and as a decimal rounded to 3 decimal places.

    log_9 (x) + 3 log_4 (x) = 14

    NOTE: This question has two different logs. What is that all about?

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  2. #2
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    use the change of base formula
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    How???

    How do I use the change-of-base formula?
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    \log_9{x} + 3\log_4{x} = 14

    \frac{\ln{x}}{\ln{9}} + \frac{3\ln{x}}{\ln{4}} = 14

    algebra time ... isolate \ln{x} and solve and calculate the value of x.
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    Quote Originally Posted by magentarita View Post
    Solve the log equation below. Express irrational solutions in exact form and as a decimal rounded to 3 decimal places.

    log_9 (x) + 3 log_4 (x) = 14

    NOTE: This question has two different logs. What is that all about?
    \log_9{x+3}\log_4{x}=14

    Change to same base (10)

    \frac{\log{x}}{\log{9}}+\frac{3\log{x}}{\log{4}}=1  4

    \frac{1}{\log{9}}\cdot\log{x}+\frac{3}{\log{4}}\cd  ot\log{x}=14

    Factor out log x,

    \log{x}\left(\frac{1}{\log{9}}+\frac{3}{\log{4}}\r  ight)=14

    \log{x}=\frac{14}{\frac{1}{\log{9}}+\frac{3}{\log{  4}}}

    Take antilog of both sides:

    x\approx209.6041493
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  6. #6
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    Thanks

    I thank you both.
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  7. #7
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    Quote Originally Posted by magentarita View Post
    Solve the log equation below. Express irrational solutions in exact form and as a decimal rounded to 3 decimal places.

    log_9 (x) + 3 log_4 (x) = 14
    ...
    Use the base-change-formula. I'll take the natural logs (see skeeter's post):

    \frac{\ln{x}}{\ln{9}} + \frac{3\ln{x}}{\ln{4}} = 14

    Solve for ln(x):

    \ln(x) = \dfrac{14}{\dfrac1{\ln(9)}+\dfrac3{\ln(4)}} = \dfrac{14}{\dfrac{\ln(4)+3\ln(9)}{\ln(9) \cdot \ln(4)}} = \dfrac{14 \cdot \ln(9) \cdot \ln(4)}{\ln(4)+3\ln(9)}

    Now use this term as exponent to the base e to get the solution exactly:

    x = e^{\frac{14 \cdot \ln(9) \cdot \ln(4)}{\ln(4)+3\ln(9)}}

    For the approximate value see master's post.
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  8. #8
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    Thank you

    Thank you earboth as well.
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