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Math Help - Logarithmic Equations

  1. #1
    Member >_<SHY_GUY>_<'s Avatar
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    Question Logarithmic Equations

    I've been trying to solve this equation for more than an hour already and I still get the wrong answer:

    log5 (x) - log25 (x) + log125 (x) = 5

    The answer is 5^6 or 15,625

    but I keep getting 5^(15/2) for some reason.
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  2. #2
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    There's some pretty nasty log bases there. Have you tried to change base?

    i.e. \log_5x = \frac{\ln x}{\ln{5}}
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  3. #3
    Member >_<SHY_GUY>_<'s Avatar
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    I tried solving the equation with base Ln, but i get lost with the fractions and I get a very ugly number, so to speak
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  4. #4
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    Maybe you can work backwards given

    \log_5 5^6 - \log_{25} 5^6 + \log_{125}5^6

    6\log_5 5 - \log_{25} (5^2)^3 + \log_{125}(5^3)^2

    6\log_5 5 - 3\log_{25} 5^2 + 2\log_{125}5^3

    6\log_5 5 - 3\log_{25} 25 + 2\log_{125}125

    6-3+2=5
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  5. #5
    Member >_<SHY_GUY>_<'s Avatar
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    I did that just to make sure the answer was correct, but I still didn't get anywhere
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  6. #6
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    \log_5 x - \log_{25} x + \log_{125} x = 5

    \frac{\ln x}{\ln 5} - \frac{\ln x}{\ln 25} + \frac{\ln x}{\ln 125} = 5

    \ln x \cdot \bigg( \frac{1}{\ln 5} - \frac{1}{\ln 25} + \frac{1}{\ln 125} \bigg)= 5

    Can you carry on from there now?
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  7. #7
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    Hello, >_<SHY_GUY>_<!

    \log_5(x) - \log_{25}(x) + \log_{125}(x) \:=\: 5

    Using the Base-change Formula, we have:

    . . . \displaystyle \frac{\ln x}{\ln 5} - \frac{\ln x}{\ln25} + \frac{\ln x}{\ln125} \;=\;5

    . . . . \displaystyle \frac{\ln x}{\ln 5} - \frac{\ln x}{\ln 5^2} + \frac{\ln x}{\ln5^3} \;=\;5

    . . . \displaystyle \frac{\ln x}{\ln 5} - \frac{\ln x}{2\ln 5} + \frac{\ln x}{3\ln 5} \;=\;5

    . . \displaystyle \frac{6\ln x - 3\ln x + 2\ln x}{6\ln 5} \;=\;5

    . . . . . . . . . . . . . . \displaystyle \frac{5\ln x}{6\ln 5} \;=\;5

    . . . . . . . . . . . . . . 5\ln x \;=\;30\ln 5

    . . . . . . . . . . . . . . . \ln x \;=\;6\ln 5

    . . . . . . . . . . . . . . .  \ln x \;=\;\ln 5^6

    . . . . . . . . . . . . . . . . x \;=\;5^6
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