Question

A^2+b^2=7ab,prove that log 1÷3(a+b)=1÷2(log a+log b)

Answer 1

Answer:

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