Question

solve the problem give me ans​

Answer 1

It is Given,

2^x = 3^y = (1/6^z)

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

(1)

2^x = k

=> 2 = k^(1/x)

(2)

3^y = k

=> 3 = k^(1/y)

(3)

6^(-z) = k

=> 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

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