Question

$if A^{x}=B^{y}=C^{z}$
and
$A^{2}=BCz=?$
it answer is
$\frac{xy}{2y-x}$

Answer 1

A^x = B^y = C^z

A^x = B^y

B= A^( x/y)

A^x = c^z

C= A^( x/z)

A^2 = BC

A^2 = A^( x/y) A^( x/z)

A^2 = A^( x/y + x/z)

compare

2 = x/y + x/z

2 - x/y = x/z

z= x/( 2 - x/y)

= x/( 2y - x)/y

= xy/( 2y-x)

Answer 2

A^x = B^y = C^z

Or A^x = B^y

A= B^x/y

B= A^x/y ...... 1

AND,

A^x = C^z

A = C^x/z

C = A^x/z ....... 2

Here,

From eq. 1 and 2,

A² = BC

A²= A^x/y * A^x/z

Since base are same.

2 = x/y * x/z

2-x/y = x/z

2y-x/y =x/z

Z= x* y/2y-x

Z= xy/2y-x

Hence the answer is verify.