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6 Aug 2009 11:50:31 IST

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a^2+b^2=7ab prove 2 log (a+b)= log 9+ log a+ log b

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a^2+b^2=7ab prove 2 log (a+b)= log 9+ log a+ log b

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hemanth ch

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Joined: 6 May 2009

Posts: 205

6 Aug 2009 11:59:06 IST

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a^2 + b^2 =7ab

add '2ab' on both sides

----> a^2 + b^2 + 2ab = 7ab + 2ab

-----> ( a+ b)^2 = 9ab

take 'log' on both sides

------> log ( a + b )^2 = log 9ab

we know that 1) log x^2 = 2 log x and

2) log xyz = log x + log y + log z

so, 2 log (a+b ) = log 9 + log a + log b

rate me if correct

Priyesh

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Joined: 18 Feb 2007

Posts: 1042

6 Aug 2009 12:01:18 IST

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L.H.S

2 log (a+b)= log(a+b)^2 = log(a^2 + b^2 + 2ab)

but a^2+b^2 = 7ab

=>log(a^2 + b^2 + 2ab) = log9ab = log9 + loga + logb (since log(a*b) = loga + logb)

=R.H.S Hence proved :)

Vijay Pal

Forum Expert

Joined: 1 Jun 2009

Posts: 2305

6 Aug 2009 12:07:12 IST

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given that

a^2+b^2=7ab ---------(1)

we know that (a+b)² =a²+b² +2ab

put the value from equation (1) we get

(a+b²)=7ab+2ab

(a+b)² =9ab

niow taking log in both the side

log (a+b)² =log(9ab)

2log(a+b)=log 9+log a+log b

hence proved

Cheers............

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