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Math Help - 3 more problems

  1. #1
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    3 more problems

    1. 2ln(3x) = 8


    2. Logx + log (x + 15) = 2

    3. Ln (x - 4) - ln (x + 1) = ln 6
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by garbles View Post
    1. 2ln(3x) = 8
    2\ln \left( {3x} \right) = 8 \Leftrightarrow \ln \left( {3x} \right) = 4 \Leftrightarrow 3x = {e^4} \Leftrightarrow x = \frac{{{e^4}}}{3}.
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  3. #3
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    Thanks just need the last 2
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  4. #4
    Senior Member DeMath's Avatar
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    Quote Originally Posted by garbles View Post

    3. Ln (x - 4) - ln (x + 1) = ln 6
    \underbrace {\ln \left( {x - 4} \right) - \ln \left( {x + 1} \right)}_{\ln a - \ln b = \ln \frac{a}{b}} = \ln 6 \Leftrightarrow \ln \frac{{x - 4}}{{x + 1}} = \ln 6 \Leftrightarrow

    \Leftrightarrow \frac{{x - 4}}{{x + 1}} = 6 \Leftrightarrow x - 4 = 6\left( {x + 1} \right) \Leftrightarrow 5x =  - 10 \Leftrightarrow x =  - 2.
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  5. #5
    Senior Member DeMath's Avatar
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    Quote Originally Posted by garbles View Post

    2. Logx + log (x + 15) = 2
    What is a basis of these logarithms?

    {\log _{\color{red}{?}}}x + {\log _{\color{red}{?}}}\left( {x + 15} \right) = 2.
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  6. #6
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    Quote Originally Posted by DeMath View Post
    What is a basis these logarithms?

    {\log _{\color{red}{?}}}x + {\log _{\color{red}{?}}}\left( {x + 15} \right) = 2.
    It doesn't say
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  7. #7
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by DeMath View Post
    What is a basis these logarithms?

    {\log _{\color{red}{?}}}x + {\log _{\color{red}{?}}}\left( {x + 15} \right) = 2.
    I'd just assume base 10.

    So, log_{10}(x(x+15))=log_{10}(x^{2}+15x)

    Thus, 10^{2}=x^{2}+15x and x^{2}+15x-100=0 becomes the equation you have to solve for.

    You will want to keep the non-negative answer.
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  8. #8
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    Quote Originally Posted by garbles View Post
    2. \Log{x}+ \log {(x + 15)} = 2

    \log{x}+\log{(x+15)}=2
    \log{(x(x+15))}=2
    \log{(x^2+15x)}=2
    x^2+15x=10^2
    x^2+15x-100=0
    (x+20)(x-5)=0
    x=5 not -20 since \log{-20} is undefined
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