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Math Help - Logarithmic Question.

  1. #1
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    Logarithmic Question.

    . Mr. Logarithmic is showing his class a solution to a problem. Is Mr. Logarithmic correct? Explain and correct if he is incorrect. Express 2log25 - log2(4/5) + 1/2log216 as a simple logarithm = 2log25 - log2(4/5) + 1/2log216 = log152-log2(4/5) + log2(16/2) = log225 - log2(4/5)+ log2(8) = log2 (25 x 4/5 x 8)


    So, from my understanding, the solution is incorrect the correct solution according to my calculations is:'

    2log25 - log2(4/5) + 1/2log216
    log225 - log2(4/5) + log24
    Therefore, we get:

    log2 (25/(4/5) X 4)

    I'm really confused here. Is my working/solution correct for the simplification of the logarithm? Any helpful tips/suggestions would be greatly appreciated.

    Thanks!
    Last edited by mr fantastic; June 5th 2010 at 05:41 PM. Reason: Edited post title.
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  2. #2
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    Quote Originally Posted by spoc21 View Post
    . Mr. Logarithmic is showing his class a solution to a problem. Is Mr. Logarithmic correct? Explain and correct if he is incorrect. Express 2log25 - log2(4/5) + 1/2log216 as a simple logarithm
    l
    your notation leaves a bit to be desired ... is this the expression?

    2\log_2{5} - \log_2\left(\frac{4}{5}\right) + \frac{1}{2}\log_2{16}<br />
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  3. #3
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    Quote Originally Posted by skeeter View Post
    your notation leaves a bit to be desired ... is this the expression?

    2\log_2{5} - \log_2\left(\frac{4}{5}\right) + \frac{1}{2}\log_2{16}<br />
    yes, thats exactly how it is.
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  4. #4
    Super Member Bacterius's Avatar
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    Didn't see the minus sign ... corrected

    Hi !
    Let's see how it goes.

    2\log_2{(5)} - \log_2{\left(\frac{4}{5}\right)} + \frac{1}{2}\log_2{(16)}

    All the logarithms have the same base, so we won't have to change this. First, simplify any coefficients in front of the logarithms, using the power law :

    \log_2{(5^2)} - \log_2\left(\frac{4}{5}\right) + \log_2{\left (16^{\frac{1}{2}} \right )}

    Now simplify the expressions :

    \log_2{(25)} - \log_2\left(\frac{4}{5}\right) + \log_2{(4)}

    Now it gets tricky : you need brackets otherwise it's not going to work. So put the last two terms into brackets :

    \log_2{(25)} - \left ( \log_2\left(\frac{4}{5}\right) - \log_2{(4)} \right )

    Now use the logarithm division/substraction law on the two last terms :

    \log_2{(25)} - \log_2{\left (\frac{\frac{4}{5}}{4} \right )}

    \log_2{(25)} - \log_2{\left (\frac{4}{20} \right )}

    Now divide using the logarithm division/substraction law :

    \log_2{\left ( \frac{25}{\frac{4}{20}} \right )}

    \log_2{(125)}

    3 \log_2{(5)}

    This is the simplest form of this expression.
    ____________________________________

    Mr. Logarithmic is wrong, and your reasoning is correct although a bit hard to follow without proper mathematical fonts. You should have simplified further to get 125 which will have concluded the exercise.
    Last edited by Bacterius; May 29th 2010 at 09:43 PM.
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  5. #5
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    Hello, spoc21!

    Mr. Logarithmic is showing his class a solution to a problem.
    Is Mr. Logarithmic correct? Explain and correct if he is incorrect.
    . . He made two errors.

    Express 2\log_2(5) - \log_2(\tfrac{4}{5}) + \tfrac{1}{2}\log_2(16) as a simple logarithm.


    2\log_2(5) - \log_2\tfrac{4}{5}) + \tfrac{1}{2}\log_2(16)

    . . =\;\log_2\left(5^2\right) -\log_2(\tfrac{4}{5}) + \log_2\left({\color{red}\tfrac{16}{2}}\right)

    . . =\; \log_2(25) - \log_2\left(\tfrac{4}{5}\right)+ \log_2(8)

    . . =\; \log_2\left(25 \,{\color{red}\times}\,  \tfrac{4}{5} \times 8\right)

    2\log_2(5) - \log_2(\tfrac{4}{5}) + \tfrac{1}{2}\log_2(16)

    . . =\;\log_2(5^2) - \bigg[\log_2(4)-\log_2(5)\bigg] + \log_2(16^{\frac{1}{2}})

    . . =\;\log_2(25) - \log_2(4) + \log_2(5) + \log_2(4)

    . . =\;\log_2(25) + \log_2(5)

    . . =\;\log_2(25 \times 5)

    . . =\;\log_2(125)

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