If a^2+b^2=79ab then prove that 2 log(a+b)=4 log3+loga+logb
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Question:-
If a^2 + b^2 = 79ab then prove that 2log(a+b) = 4log3+loga+logb
Note:-
- (a+b)^2 = a^2 + b^2 + 2ab
- (a-b)^2 = a^2 + b^2 - 2ab
- log(abc) = loga + logb + logc
- log(a/b) = loga - logb
- log(a^m) = mloga
Solution:-
- Considering the given statement ,a^2 + b^2 = 79ab
- a^2 + b^2 = 79ab
- Adding 2ab on both sides
- a^2 + b^2 + 2ab = 79ab + 2ab
- (a+b)^2 = 81ab
- Applying log on both sides,
- log ( (a+b)^2 ) = log (81ab)
- 2log(a+b) = log(3^4 . a . b)
- 2log(a+b) = log(3^4) + loga + logb
- 2log(a+b) = 4log3 + loga + logb
Hence ,
If a^2 + b^2 = 79ab, then 2log(a+b) = 4log3+loga+logb
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