A^2+b^2=7ab,prove that log 1÷3(a+b)=1÷2(log a+log b)
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a^2+b^2=7ab
we add both side 2ab
a^2+b^2+2ab= 7ab+2ab
(a+b)^2=9ab
apply log on both side
log(a+b)^2=log9ab
2log(a+b)=log9+log a+ log b
2log (a+b)= log 3^2+ log a +log b
2log (a+b)= 2log3 + log a +log b
2log(a+b)-2log 3= log a+ log b
2log (a+b)= log a+log b
_____
3
log(a+b)/3=log a+ log b
____________
2
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