Question

solve it with more steps plzz​

Answer 1

Hey Mate here is your answer ✌️☺️

Answer 2

2^x=3^y=6^-z = k (let)

k=2^x

or (k)^1/x=2

similary (k)^1/y=3

(k)^-1/z=6

or (k)^-1/z=2×3=(k)^1/x × (k)^1/y

(k)^-1/z =(k)^1/x+1/y

Hence,

-1/z =1/x + 1/y

0=1/x +1/y +1/z ans