Question

if 2^x=3^y=6^-z then find 1/x +1/y+1/z

Answer 1

Answér :

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

Solution :

• Given : 2^x = 3^y = 6^(-z)
• To find : 1/x + 1/y + 1/z = ?

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

• If 2^x = k

then 2 = k^(1/x)

• If 2^y = k

then 3 = k^(1/y)

• If 6^(-z) = k

then 6 = k^(1/-z)

=> 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)

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

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

Hence ,

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

Answer 2

Thus 1/z will be equal to - log 6 to base k. Given 2^x=3^y=6^-z. Thus 1/z will be equal to - log 6 to base k.