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Math Help - Solving Logarithmic Equations

  1. #1
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    Solving Logarithmic Equations

    I need help trying to figure out how to solve

    log (2-3x) + log (3-2x) = 1.5

    logbase7(1-x) - logbase7(x+2) = logbase7(x^2)

    Thank You!
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  2. #2
    Moo
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    Hello,

    log (2-3x) + log (3-2x) = 1.5
    You know that log(a)+log(b)=log(ab)

    So \log(2-3x)+\log(3-2x)=\log((2-3x)(3-2x))=1.5

    And you can continue by developping what's in brackets and taking the exponential (beware of the domain of the function)

    logbase7(1-x) - logbase7(x+2) = logbase7(x^2)
    Since \log_{x} y = \frac{\log(y)}{\log{x}}, you can transform the previous equation :

    \log_7 (1-x) - \log_7 (x+2) = log_7 (x^2)  \Longleftrightarrow \log (1-x) - \log(x+2) = \log(x^2)

    Since log(a)-log(b)=log(a/b), it's the same as :

    \log(\frac{1-x}{x+2}) = \log(x^2)

    And if log(a)=log(b), then a=b
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  3. #3
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    Quote Originally Posted by Moo View Post
    Hello,



    You know that log(a)+log(b)=log(ab)

    So \log(2-3x)+\log(3-2x)=\log((2-3x)(3-2x))=1.5

    And you can continue by developping what's in brackets and taking the exponential (beware of the domain of the function)


    I don't know what to do after that. (log((2-3x)(3-2x))=1.5) am I supposed to subtract 1.5 and get 6x^2- 13x + 4.5? And go from there?
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  4. #4
    Moo
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    Nope,

    You have log(....)=1.5

    So "...." = exp(1.5)
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  5. #5
    Senior Member Twig's Avatar
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    Correct?

    Hi!

    Is this correctly simplified?

    log (6-13x+6x^2) = 1.5 then..

    6x^2 - 13x + 6 = 10^1.5

    then solve this second degree equation..

    I get x1 = 3,41 x2= -1,25
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