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Math Help - Logs :(

  1. #1
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    Logs :(

    Ok, I'm having some issues with these questions. So could you show me step by step on how to come to an answer.

    1. write as a single logarithm in simplest form
    (Log 16x^8)/4 - (Log 27x)/3

    2. isolate x using laws of logarithims
    3 Logx(4)=2
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  2. #2
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    Re: Logs :(

    Quote Originally Posted by Twinkie View Post
    1. write as a single logarithm in simplest form
    (Log 16x^8)/4 - (Log 27x)/3
    I'm afraid that I don't understand the final goal: "simplest form." So I'll show you how I would do the problem and you can make your own decision if the answer is simple enough.

    We're going to be using the change of base formula:
    \log_a(x) = \frac{\log_b(x)}{\log_b(a)}

    I'm going to choose my favorite base, e.

    So
    \frac{1}{4}\log_{16}(x^8) - \frac{1}{3} \log_{27}(x)

    Skipping a couple of steps, you should arrive at
    2 \left [ \frac{ln(x)}{ln(16)} \right ] - \frac{1}{3} \left [ \frac{ln(x)}{ln(27)} \right ]

    Which has the final form, as simply as I can write it:
    \left [ \frac{1}{2~ln(2)} - \frac{1}{9~ln(3)} \right ] ln(x)

    The only other thing I can think of would be to subtract the fractions.

    -Dan
    Last edited by topsquark; December 10th 2012 at 05:28 PM. Reason: Made an oopsie
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  3. #3
    Forum Admin topsquark's Avatar
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    Re: Logs :(

    Quote Originally Posted by Twinkie View Post
    2. isolate x using laws of logarithims
    3 Logx(4)=2
    \log_x(4) = 2

    x^{\log_x(4)} = x^2

    4 = x^2

    And you can finish from there. Watch out for the extra (wrong) solution that comes from this.

    -Dan
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  4. #4
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    Re: Logs :(

    Ok thankyou for your help.
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  5. #5
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    Re: Logs :(

    1. log 16x^8/4 - log27x/3
    2.log4x^8 - log 9x
    3.log4x^8/9x
    4.log 4/9x^7
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  6. #6
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    Re: Logs :(

    That kind of question makes sence now, but I have anouther question that I'm not sure how to do.
    Solve.
    3log(6) X= log(6) 9+ log(6) 24
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  7. #7
    Forum Admin topsquark's Avatar
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    Re: Logs :(

    Quote Originally Posted by bjhopper View Post
    1. log 16x^8/4 - log27x/3
    2.log4x^8 - log 9x
    The division of 4 in the first term is outside the logarithm. So you can't divide through.

    -Dan
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  8. #8
    Forum Admin topsquark's Avatar
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    Re: Logs :(

    Quote Originally Posted by Twinkie View Post
    That kind of question makes sence now, but I have anouther question that I'm not sure how to do.
    Solve.
    3log(6) X= log(6) 9+ log(6) 24
    If the bases on all the logs are the same then we can do this rather easily. Let's get everything to one side:
    2\log_6(x) - \log_6(9) - \log_6(24) = 0

    Let's rewrite that first term:
    \log_6(x^2) - \log_6(9) - \log_6(24) = 0

    Now we can use
    \log_a(p) + \log_a(q) = \log_a(pq)

    and
    \log_a(p) - \log_a(q) = \log_a \left ( \frac{p}{q} \right )

    (Note: These are only true when the base a is the same on all log functions.)

    Can you finish from here?

    -Dan
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  9. #9
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    Re: Logs :(

    Quote Originally Posted by Twinkie View Post
    That kind of question makes sence now, but I have anouther question that I'm not sure how to do.
    Solve.
    3log(6) X= log(6) 9+ log(6) 24
    Twinkie,
    1.log base 6 (x)^3= log b6 (9*24) =log b6 (216)
    2. x^3 = 216
    3. x=6
    Last edited by bjhopper; December 11th 2012 at 03:49 AM.
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  10. #10
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    Re: Logs :(

    Quote Originally Posted by topsquark View Post
    The division of 4 in the first term is outside the logarithm. So you can't divide through.

    -Dan
    Dan, That was a mistake but when I redo the question I come up with log 2/3 X^5/3 as the simplified log
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  11. #11
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    Re: Logs :(

    Ok thanks!
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