Question

if 2^x=3^y=6^-z,1/x+1/y+1/z​

Answer 1

Answer:

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

x=logk base 2 =>1/x= log2 base k

y=logk base 3 =>1/y=log3 base k

z=-logk base6 =>1/z=-log6 base k

1/x+1/y+1/z=log (2×3) - log (6) All base k

=>log (6/6) base k

=>0 ans

Answer 2

Answer:

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

So,

2=k^1/x

3=k^1/y

6= k^-1/z

2×3 = k^-1/z

k^1/x × k^1/y = k^-1/z

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

Comparing powers as base on LHS and RHS is same.

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

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