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Math Help - One logarithm question

  1. #1
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    Exclamation One logarithm question*EDITED

    Hi, I am having problem with one difficult looking logarithm. Please help me!
    <br />
3log_2 \frac{9}{8} + 4*log_2 \frac{16}{15} - 2 log_2 \frac{24}{25}<br />

    Thanks! Please show your working -

    *EDIT: This is supposed to be solved without an calculator, so can you please simplify it down to something which you can easily work with, with logarithm laws.

    Thanks
    Last edited by BG5965; November 15th 2008 at 02:52 PM.
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  2. #2
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    Hey

    Quote Originally Posted by BG5965 View Post
    Hi, I am having problem with one difficult looking logarithm. Please help me!


    Thanks! Please show your working.
    this is really hard to read, man

     3log_2 \frac{9}{8} + 4*log_2 \frac{16}{15} - 2 log_2 \frac{24}{25}

    it is  log_a(b) = ln(b) /ln (a)

    therefor

     1/ln(2) [ 3 ln(\frac{9}{8}) + 4*ln(\frac{16}{15}) - 2 ln(\frac{24}{25})] = 0.562
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  3. #3
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    Thanks, but I wouldn't be able to do that without an calculator. Is there another way to simplify it down to basic logarithms that we can easily use?
    Thanks
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  4. #4
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    Quote Originally Posted by BG5965 View Post
    Hi, I am having problem with one difficult looking logarithm. Please help me!
    <br />
3log_2 \frac{9}{8} + 4*log_2 \frac{16}{15} - 2 log_2 \frac{24}{25}<br />

    Thanks! Please show your working -

    *EDIT: This is supposed to be solved without an calculator, so can you please simplify it down to something which you can easily work with, with logarithm laws.

    Thanks
    From the usual log rules:

    = 3 (\log_2 9 - \log_2 8) + 4 (\log_2 16 - \log_2 15) - 2 (\log_2 24 - \log_2 25)

    = 3 (\log_2 3^2 - \log_2 2^3) + 4 (\log_2 2^4 - \log_2 (3 \cdot 5)) - 2 (\log_2 (3 \cdot 2^3) - \log_2 5^2)

    From the usual log rules:

    = 3 (2 \log_2 3 - 3) + 4 (4 - (\log_2 3 + \log_2 5) ) - 2 (\log_2 3 + \log_2 2^3 - 2 \log_2 5)

    = 3 (2 \log_2 3 - 3) + 4 (4 - \log_2 3 - \log_2 5 ) - 2 (\log_2 3 + 3 - 2 \log_2 5)

    = 6 \log_2 3 - 9 + 16 - 4 \log_2 3 - 4 \log_2 5 - 2 \log_2 3 -6 + 4 \log_2 5


    and the mopping up is left for you to do.

    I get 1 as the answer.


    But you should check all my work carefully because I am well known for making careless mistakes.
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  5. #5
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    Thumbs up

    Thanks - I just realized logarithms can sometime be just 'complex' algebra.
    Anyway, I canceled out all the like terms (the logarithms with the common base) and was left with -9 + 10 = 1

    I think you were right.

    Thanks for your help.
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  6. #6
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    Hello, BG5965!

    No wonder you're having problems . . . this is a messy one!

    And I agree with Mr. F . . .


    3\log_2\left(\tfrac{9}{8}\right) + 4\log_2\left(\tfrac{16}{15}\right) - 2\log_2\left(\tfrac{24}{25}\right)
    I'll work on the three terms separately, and combine them at the end.


    3\log_2\left(\frac{9}{8}\right) \;=\;3\bigg[\log_2(9) - \log_2(8)\bigg] \;=\;3\bigg[\log_2(3^2) - \log_2(2^3)\bigg]

    . . =\;3\bigg[2\log_2(3) - 3\bigg] \;=\;\boxed{6\log_2(3) - 9}


    4\log_2\left(\frac{16}{15}\right) \;=\;4\bigg[\log_2(16) - \log_2(15)\bigg] \;=\;4\bigg[\log_2(2^4) - \log_2(3\cdot5)\bigg]

    . . = \;4\bigg[4 - \bigg(\log_2(3) + \log_2(5)\bigg)\bigg] \;=\;\boxed{16 - 4\log_2(3) - 4\log_2(5)}


    -2\log_2\left(\frac{24}{25}\right) \;=\;-2\bigg[\log_2(24) - \log_2(25)\bigg] \;=\;-2\bigg[\log_2(2^3\cdot3) - \log_2(5^2)\bigg]

    . . = \;-2\bigg[\log_2(2^3) + \log_2(3) - 2\log_2(5)\bigg] \;=\;-2\bigg[3 +  \log_2(3) - 2\log_2(5)\bigg]

    . . = \;\boxed{-6 - 2\log_2(3) + 4\log_2(5)}



    Combine the three expressions:

    . . \begin{array}{ccccccc}6\log_2(3) & & &  - & 9 \\<br />
\text{-}4\log_2(3) & - & 4\log_2(5) & + & 16 \\ \text{-}2\log_2(3) &+& 4\log_2(5) &-& 6 \\ \hline \\[-4mm]<br />
& & & &\boxed{{\color{red}1}} \end{array}

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