Real numbers y, p, and n (all greater than 1) satisfy

$$\left({\log }_{p^{1/n}}y\right)\left({\log }_{y^{1/n}}p\right)=16,$$ $\text{where the logarithms are taken to the bases}{p}^{1/n}\text{and}{y}^{1/n}.$ $\text{The value of }\mathrm{n }\text{is --------}$

The correct option is 4.

Let's solve the problem step-by-step to understand why this option is correct.

We are given that the real numbers $y$, $p$, and $n$ (all greater than 1) satisfy the equation:

$$\left({\log }_{p^{1/n}}y\right)\left({\log }_{y^{1/n}}p\right)=16$$

To simplify the expressions, we can use the change-of-base property of logarithms, which states that for any base $b$ and argument $a$, we have ${\log }_{b}a=\frac{\ln a}{\ln b}$ (using natural logarithm, or log to any common base).

Applying this to the first term:

$${\log }_{p^{1/n}}y=\frac{\ln y}{\ln \left(p^{1/n}\right)}$$

Using the property of logarithms where $\ln \left(x^k\right)=k\ln x$, we get:

$${\log }_{p^{1/n}}y=\frac{\ln y}{\frac{1}{n}\ln p}=n\frac{\ln y}{\ln p}$$

Similarly, applying the same properties to the second term:

$${\log }_{y^{1/n}}p=\frac{\ln p}{\ln \left(y^{1/n}\right)}=\frac{\ln p}{\frac{1}{n}\ln y}=n\frac{\ln p}{\ln y}$$

Now, let's substitute these simplified expressions back into the original equation:

$$\left(n\frac{\ln y}{\ln p}\right)·\left(n\frac{\ln p}{\ln y}\right)=16$$

Observe that the logarithmic terms cancel each other out:

$$n^2·(\frac{\ln y}{\ln p}·\frac{\ln p}{\ln y})=16$$

$$n^2·1=16$$

This leaves us with the equation:

$$n^2=16$$

Taking the square root on both sides:

$$n=\pm 4$$

Since the problem specifies that $n$ is greater than 1, we discard the negative root ($n=-4$).

Thus, we obtain:

$$n=4$$