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Math Help - 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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    \frac{1}{20}(\frac{23}{20})^{x}=5

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

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

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

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

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

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

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

    \approx{32.95}



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

     <br />
\begin{aligned}<br />
\frac{1}<br />
{{20}}\left( {\frac{{23}}<br />
{{20}}} \right)^x &= 5\\<br />
20 \cdot \frac{1}<br />
{{20}}\left( {\frac{{23}}<br />
{{20}}} \right)^x &= 20 \cdot 5\\<br />
\left( {\frac{{23}}<br />
{{20}}} \right)^x &= 100<br />
\end{aligned}<br />
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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

     <br />
\begin{aligned}<br />
6^{x + 1} &= 4^{2x - 1}\\<br />
(x + 1)\log 6 &= 2(2x - 1)\log 2\\<br />
x\log 6 + \log 6 &= 4x\log 2 - 2\log 2\\<br />
4x\log 2 - x\log 6 &= 2\log 2 + \log 6\\<br />
x(4\log 2 - \log 6) &= 2\log 2 + \log 6\\<br />
x &= \frac{{2\log 2 + \log 6}}<br />
{{4\log 2 - \log 6}}<br />
\end{aligned}<br />
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