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September 19th, 2014, 08:36 AM  #1 
Newbie Joined: Sep 2014 From: india Posts: 1 Thanks: 0  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. 
September 19th, 2014, 09:13 AM  #2 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,157 Thanks: 732 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
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. 
September 19th, 2014, 02:54 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,929 Thanks: 2205 
7log(10/9)  2log(25/24) + 3log(81/80) = log(2)


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