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Math Help - Is it possible to not use a calculator and solve this problem?

  1. #1
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    Is it possible to not use a calculator and solve this problem?

    Log(8)+3/5Log(14) - 3/5Log(7)+Log(1/(213/5))

    i got log (8*143/5) / Log (73/5 *1/(213/5 )) = Log(2)

    Just wondering if you can solve this without using a calculator since my exams forbid calculators.
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  2. #2
    Super Member Matt Westwood's Avatar
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    Re: Is it possible to not use a calculator and solve this problem?

    Quote Originally Posted by chewydrop View Post
    Log(8)+3/5Log(14) - 3/5Log(7)+Log(1/(213/5))

    i got log (8*143/5) / Log (73/5 *1/(213/5 )) = Log(2)

    Just wondering if you can solve this without using a calculator since my exams forbid calculators.
    You mean you *did* use a calculator? Where??
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  3. #3
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    Re: Is it possible to not use a calculator and solve this problem?

    Quote Originally Posted by Matt Westwood View Post
    You mean you *did* use a calculator? Where??
    (8*143/5) / (73/5 *1/(213/5 )) into the calculator and got 2.0000001
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  4. #4
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    Re: Is it possible to not use a calculator and solve this problem?

    This is a simple exercise in properties of the exponential. Do you not know that log(a^x)= x log(a) and log(ab)= log(a)+ log(b)?

    [QUOTE=chewydrop;830170]Log(8)+3/5Log(14) - 3/5Log(7)+Log(1/(213/5))[/tex]
    8= 2^3 so [tex]log(8)= 3 log(2). 14= 2(7) so (3/5)log(14)= (3/5)log(2)+ (3/5)log(7). 1/2^{13/5}= 2^{-13}{5} so log(1/2^{13/5})= -(13/5)log(2)

    So you have 3 log(2)+ (3/5)log(2)+ (3/5)log(7)- (3/5)log(7)- 13/5 log(2)

    Now, it's just arithmetic.

    i got log (8*143/5) / Log (73/5 *1/(213/5 )) = Log(2)

    Just wondering if you can solve this without using a calculator since my exams forbid calculators.
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    Re: Is it possible to not use a calculator and solve this problem?

    Hello, chewydrop!

    \text{Simplify: }\:\log(8)+\tfrac{3}{5}\log(14) - \tfrac{3}{5}\log(7)+\log\left(\frac{1}{2^{\frac{13  }{5}}}\right)

    Re-group terms: . \left[\log(8) + \log\left(2^{-\frac{13}{5}}\right)\right] + \left[\tfrac{3}{5}\log(14) - \tfrac{3}{5}\log(7)\right]

    . . . =\;\left[\log(2^3) + \log(2^{-\frac{13}{5}})\right] + \tfrac{3}{5}\left[\log(14) - \log(7)\right]

    . . . =\; \log\left(2^3\cdot2^{-\frac{13}{5}}\right) + \tfrac{3}{5}\log\left(\frac{14}{7}\right)

    . . . =\;\log\left(2^{\frac{2}{5}}\right) + \tfrac{3}{5}\log(2)


    . . . =\; \tfrac{2}{5}\log(2) + \tfrac{3}{5}\log(2)

    . . . =\; \log(2)
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    Re: Is it possible to not use a calculator and solve this problem?

    Quote Originally Posted by Soroban View Post
    Hello, chewydrop!


    Re-group terms: . \left[\log(8) + \log\left(2^{-\frac{13}{5}}\right)\right] + \left[\tfrac{3}{5}\log(14) - \tfrac{3}{5}\log(7)\right]

    . . . =\;\left[\log(2^3) + \log(2^{-\frac{13}{5}})\right] + \tfrac{3}{5}\left[\log(14) - \log(7)\right]

    . . . =\; \log\left(2^3\cdot2^{-\frac{13}{5}}\right) + \tfrac{3}{5}\log\left(\frac{14}{7}\right)

    . . . =\;\log\left(2^{\frac{2}{5}}\right) + \tfrac{3}{5}\log(2)


    . . . =\; \tfrac{2}{5}\log(2) + \tfrac{3}{5}\log(2)

    . . . =\; \log(2)
    For some reason when i regrouped it on my own i got a different expression:
    Log(8)+Log(2-13/5)-3/5Log(7)+3/5Log(14)

    would this still work? because im stuck already
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