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September 19th, 2014, 08:36 AM   #1
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Unhappy problem in logarithms

Hello every one,
I have a problem to solve following logarithms
prove that 7 Log(20/9)-2 Log(25/24)-3 Log(81/80)= Log(2)
I will get 8 Log(2) instead of Log(2)
Would you plz help me to solve this?

Last edited by skipjack; September 19th, 2014 at 02:47 PM.
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September 19th, 2014, 09:13 AM   #2
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I get $\displaystyle 2 \log \left(\frac{5^3 \times 2^{16}}{3^{12}}\right)$

However, if your question is

Work out $\displaystyle 7 \log \left(\frac{20}{9}\right) - 2 \log \left(\frac{24}{25}\right)$ + $\displaystyle 3 \log \left(\frac{81}{80}\right)$,

i.e. with a plus sign on the third term, I get $\displaystyle 8\log 2$.

If that change in sign doesn't explain it, you probably accidentally added some powers on the 2 rather than multiplying them when reducing down the $\displaystyle 24^2, 20^7$ or $\displaystyle 80^3$ parts.
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September 19th, 2014, 02:54 PM   #3
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7log(10/9) - 2log(25/24) + 3log(81/80) = log(2)
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