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Math Help - Logarithmic Equations

  1. #1
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    Logarithmic Equations

    Solve:
    log_8(x-3) - log_8 (3x + 11) = log_8 (\frac{1}{x+1})

    I have this:

    log_8(\frac {x-3}{3x + 11}) = log_8 (\frac{1}{x+1})

    log_8(-2x -14) = log_8 (-x)

    But I'm not sure my second step is right...how would you solve this equation?

    Answer: 7
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  2. #2
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    \begin{array}{l}<br />
 \log _8 \left( {x - 3} \right) - \log _8 \left( {3x + 11} \right) = \log _8 \left( {\frac{1}{{x + 1}}} \right) =  - \log _8 \left( {x + 1} \right) \\ <br />
 \log _8 \left( {\frac{{\left( {x - 3} \right)\left( {x + 1} \right)}}{{3x + 11}}} \right) = 0 \\ <br />
 \end{array}<br />
    Last edited by Plato; December 12th 2007 at 03:22 PM.
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  3. #3
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    Quote Originally Posted by Plato View Post
    \begin{array}{l}<br />
 \log _8 \left( {x - 3} \right) - \log _8 \left( {3x + 1} \right) = \log _8 \left( {\frac{1}{{x + 1}}} \right) =  - \log _8 \left( {x + 1} \right) \\ <br />
 \log _8 \left( {\frac{{\left( {x - 3} \right)\left( {x + 1} \right)}}{{3x + 1}}} \right) = 0 \\ <br />
 \end{array}<br />
    How did you get a -log, if you were to move the numerator of 1, wouldn't it be just log?
    Also, what happened to the 11 in the denominator of the first log?
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  4. #4
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    Quote Originally Posted by Macleef View Post
    what happened to the 11 in the denominator of the first log?
    That was a mere typo! I have corrected it.
    In any event it would not change the basic idea. You really should know the basic properties.

    Quote Originally Posted by Macleef View Post
    How did you get a -log, if you were to move the numerator of 1, wouldn't it be just log?
    Again you should know basic properties.
    \log _b \left( {\frac{1}{a}} \right) = \log _b \left( {a^{ - 1} } \right) =  - \log _b \left( a \right).
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  5. #5
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    Could you show me how to solve this equation step by step?
    I don't see how log_8 (\frac {1}{x+1}) = -log_8 (x+1)
    As for the negative exponent law, did you take the exponent of -1 from (x+1), is that what you did?

    And how would you solve for x in:

     (\frac {x^{2} - 2x - 3}{3x + 11}) = 0
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  6. #6
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    Quote Originally Posted by Macleef View Post
    Could you show me how to solve this equation step by step?
    I don't see how log_8 (\frac {1}{x+1}) = -log_8 (x+1)
    As for the negative exponent law, did you take the exponent of -1 from (x+1), is that what you did?
    I think that you need more help than you can expect from here.
    Go talk to your instructor. You need live one-on-one help.
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  7. #7
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    No, I don't need extra help.
    I got the question, I forgot to cross-multiply in my way.
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  8. #8
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    Quote Originally Posted by Macleef View Post
    Solve:
    log_8(x-3) - log_8 (3x + 11) = log_8 (\frac{1}{x+1})

    ...
    Hello,

    1. I agree with Plato that you only have very basic knowledge about powers and logarithms. Therefore you have to learn by heart until tomorrow the content of the attachment.

    2.
    log_8(x-3) - log_8 (3x + 11) = log_8 (\frac{1}{x+1})~\iff~  \log_8\left(\frac{x-3}{3x+11}\right) = \log_8(1) - \log_8(x+1)~,~x > 3

    Now de-logarithmize(?):

    \frac{x-3}{3x+11} = \frac{1}{x+1}~\iff~(x-3)(x+1)=3x+11 ~\iff~ x^2-5x-14=0

    Solve this quadratic equation for x. I've got: x = -2~\vee~x = 7

    x = -2 doesn't belong to the domain of the equation thus you have the solution x = 7.

    So you are right - but I really would like to know how you got this solution.
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