Question

Hola !!
Good Evening
Can you help me with this problem mates.

Thank you​

Answer 1

Let 2^x=3^y=6^-z=k., then,

2=k^(1/x)

3=k^(1/y)

6=k^(-1/z)

we know that 2×3=6

here, k^(1/x)×k^(1/y)=k^(-1/z)

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

bases are equal hence powers are equal

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

hence, 1/x+1/y+1/z=0