Question

2^x=3^y=6^-z prove 1/x+1/y+1/z = 0

Answer 1

let 2^x=3^y=6^-z=K

Therefore x log 2= y log 3 = -z log 6= K

we can write x = k/log 2

                     y = k/log 3

                     z = -k/log 6=-k/log 2+log 3

According to question

1/x+1/y+1/z

(log 2 + log 3 - log 2 - log 3)/k

0/k = 0