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Math Help - Solving a Log Equation

  1. #1
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    Solving a Log Equation

    log6(x+1) + log6(x+2) = 1

    The 6 in both equations is the base.

    I know that...

    log6(x+1) + log6(x+2) = log6(x+1)(x+2)

    So I get to here.

    log6(x+1)(x+2) = 1

    I assume we have to make it equal to 0.

    log6(x^2+3x+2) = 1
    log6(x^2+3x+1) = 0

    Tell me if I'm doing something wrong here.

    The answer is supposed to be 1, but I don't understand how to get to the answer.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jeavus View Post
    log6(x+1) + log6(x+2) = 1

    The 6 in both equations is the base.

    I know that...

    log6(x+1) + log6(x+2) = log6(x+1)(x+2)

    So I get to here.

    log6(x+1)(x+2) = 1

    I assume we have to make it equal to 0.

    log6(x^2+3x+2) = 1
    log6(x^2+3x+1) = 0
    that's wrong:

    remember the definition of a logarithm... If \log_a b = c then a^c = b

    so you have: \log_6 (x^2 + 3x + 2) = 1

    \Rightarrow 6^1 = x^2 + 3x + 2

    now continue
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  3. #3
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    Thanks again.
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  4. #4
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    How do I go about solving this equation?

    log5(2x+2) - log5(x-1) = log5(x+1)

    The 5 in each is the base.
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  5. #5
    MHF Contributor red_dog's Avatar
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    Use the formula \displaystyle\log_ab-\log_ac=\log_a\frac{b}{c}.
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  6. #6
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    Hmmm...

    log5(2x+2) - log5(x-1) = log5(x+1)
    log5(2x+2/x-1) = log5(x+1)
    log5(2x+2/x-1)(x+1) = 0

    Where do I go from here?
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jeavus View Post
    Hmmm...

    log5(2x+2) - log5(x-1) = log5(x+1)
    log5(2x+2/x-1) = log5(x+1)
    log5(2x+2/x-1)(x+1) = 0

    Where do I go from here?
    how did you get \log_5 \left( \frac {2x + 2}{x - 1} \right)(x + 1) exactly? you had to subtract \log_5 (x + 1) correct?

    when you find what the log should be, follow what i told you in the first post


    ...or, you could have simply dropped all the logs...

    \log_5 \left( \frac {2x + 2}{x - 1}\right) = \log_5 (x + 1)

    \Rightarrow \frac {2x + 2}{x - 1} = x + 1 ............do you see why?
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  8. #8
    MHF Contributor red_dog's Avatar
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    \displaystyle\log_5\frac{2x+2}{x-1}=\log_5(x+1)\Rightarrow\frac{2x+2}{x-1}=x-1
    Now, solve for x.
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  9. #9
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    Thanks very much.
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