If \( 3^{x}=5 ; 5^{y}=7 \) and \( 7^{z}=9 \). Calculate the value of \( x y z \)

To solve for \( xyz \) given the equations \( 3^{x}=5 \), \( 5^{y}=7 \), and \( 7^{z}=9 \), we start by expressing \( x \), \( y \), and \( z \) in terms of logarithms. 1. From \( 3^x = 5 \), we can take the logarithm of both sides: \[ x = \log_3(5) = \frac{\log(5)}{\log(3)} \] 2. From \( 5^y = 7 \): \[ y = \log_5(7) = \frac{\log(7)}{\log(5)} \] 3. From \( 7^z = 9 \): \[ z = \log_7(9) = \frac{\log(9)}{\log(7)} = \frac{2\log(3)}{\log(7)} \quad \text{(since } \log(9) = \log(3^2) = 2\log(3)\text{)} \] Now, we multiply \( xyz \): \[ xyz = \left(\frac{\log(5)}{\log(3)}\right) \left(\frac{\log(7)}{\log(5)}\right) \left(\frac{2\log(3)}{\log(7)}\right) \] Notice that in our product, \( \log(5) \) in the numerator of \( y \) cancels with \( \log(5) \) in the denominator of \( x \), and \( \log(7) \) does the same: \[ xyz = \frac{\log(3)}{\log(3)} \cdot 2 = 2 \] Therefore, the value of \( xyz \) is: \[ \boxed{2} \]