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Math Help - Logarithms

  1. #1
    Junior Member
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    Johor Bahru
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    Logarithms

    Solve each of the following eq.

    1. 2 log(base 2)(x + 2) = log(base2)(x + 14)

    2. log(base 27)(x) = log(base x)(27)

    How to start?
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Remember that you can remove the bases once there are only logs. So, raise the power

    2log_2(x+2) = \log_2(x+14)

    log_2(x+2)^2 = \log_2(x+14)

    (x+2)^2 = x + 14

    For the other one, change the base.

    log_a b = \dfrac{1}{\log_b a}

    EDIT: I didn't notice the '2'. Changed it now.
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  3. #3
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    e^(i*pi)'s Avatar
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    2\log_2(x+2) = \log_2[(x+2)^2]

    In other words (x+2)^2 = x+14 - solve the quadratic but remember that x must be greater than -2


    For number 2 use the change of base rule

    \dfrac{\ln(x)}{\ln(27)} = \dfrac{\ln(27)}{\ln(x)} \rightarrow \: [\ln(x)]^2 - [\ln(27)]^2 = 0

    Solve using the difference of two squares. Again, from the domain, x > 0 \text { and  } x \neq 1
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  4. #4
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    For no.2

    After 1/log(base x)(27) =log(base x)(27)

    Then,I'm kinda confused.....Is this the correct way?
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  5. #5
    Senior Member Educated's Avatar
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    \dfrac{1}{\log_x(27)} = \log_x (27)

    Yes that is correct.

    Now bring the x's to one side giving us:

    1 = (\log_x (27))^2

    \sqrt{1} = \log_x (27)

    Can you carry on from there?
    (Remember \sqrt{1} gives 2 answers)
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