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Thread: Exponential and Logarithmic problems

  1. #1
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    Exponential and Logarithmic problems

    I can do such problems like

    6^x+1 = 4^2x-1

    but I can't figure out how to do

    .05(1.15)^X = 5



    Help =(
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  2. #2
    Eater of Worlds
    galactus's Avatar
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    $\displaystyle \frac{1}{20}(\frac{23}{20})^{x}=5$

    $\displaystyle (\frac{23}{20})^{x}=100$

    $\displaystyle xln(\frac{23}{20})=ln(100)$

    $\displaystyle x=\frac{2ln(10)}{ln(\frac{23}{20})}$

    $\displaystyle \approx{32.95}$
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  3. #3
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    Quote Originally Posted by galactus View Post
    $\displaystyle \frac{1}{20}(\frac{23}{20})^{x}=5$

    $\displaystyle (\frac{23}{20})^{x}=100$

    $\displaystyle xln(\frac{23}{20})=ln(100)$

    $\displaystyle x=\frac{2ln(10)}{ln(\frac{23}{20})}$

    $\displaystyle \approx{32.95}$



    Where did the 100 on the other side come from?
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  4. #4
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    Krizalid's Avatar
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    It's really simple

    $\displaystyle
    \begin{aligned}
    \frac{1}
    {{20}}\left( {\frac{{23}}
    {{20}}} \right)^x &= 5\\
    20 \cdot \frac{1}
    {{20}}\left( {\frac{{23}}
    {{20}}} \right)^x &= 20 \cdot 5\\
    \left( {\frac{{23}}
    {{20}}} \right)^x &= 100
    \end{aligned}
    $
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  5. #5
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by JonathanEyoon View Post
    6^x+1 = 4^2x-1
    I assume you forgot the brackets

    $\displaystyle
    \begin{aligned}
    6^{x + 1} &= 4^{2x - 1}\\
    (x + 1)\log 6 &= 2(2x - 1)\log 2\\
    x\log 6 + \log 6 &= 4x\log 2 - 2\log 2\\
    4x\log 2 - x\log 6 &= 2\log 2 + \log 6\\
    x(4\log 2 - \log 6) &= 2\log 2 + \log 6\\
    x &= \frac{{2\log 2 + \log 6}}
    {{4\log 2 - \log 6}}
    \end{aligned}
    $
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