Question

If $\frac{{\left(p+{\displaystyle \frac{1}{q}}\right)}^{p-q}{\left(p-{\displaystyle \frac{1}{q}}\right)}^{p+q}}{{\left(q+{\displaystyle \frac{1}{p}}\right)}^{p-q}{\left(q-{\displaystyle \frac{1}{p}}\right)}^{p+q}}={\left(\frac{p}{q}\right)}^x,\mathrm{then}x=\_\_\_\_\_\_\_\_\_.$

Answer 1

$$\begin{gathered} \frac{{\left(p+{\displaystyle \frac{1}{q}}\right)}^{p-q}{\left(p-{\displaystyle \frac{1}{q}}\right)}^{p+q}}{{\left(q+{\displaystyle \frac{1}{p}}\right)}^{p-q}{\left(q-{\displaystyle \frac{1}{p}}\right)}^{p+q}}={\left(\frac{p}{q}\right)}^x \\ \Rightarrow \frac{{\left({\displaystyle \frac{qp+1}{q}}\right)}^{p-q}{\left({\displaystyle \frac{pq-1}{q}}\right)}^{p+q}}{{\left({\displaystyle \frac{pq+1}{p}}\right)}^{p-q}{\left({\displaystyle \frac{pq-1}{p}}\right)}^{p+q}}={\left(\frac{p}{q}\right)}^x \\ \Rightarrow \frac{{\displaystyle \frac{{\left(qp+1\right)}^{p-q}}{{\left(q\right)}^{p-q}}}\times {\displaystyle \frac{{\left(pq-1\right)}^{p+q}}{{\left(q\right)}^{p+q}}}}{{\displaystyle \frac{{\left(pq+1\right)}^{p-q}}{{\left(p\right)}^{p-q}}}\times {\displaystyle \frac{{\left(pq-1\right)}^{p+q}}{{\left(p\right)}^{p+q}}}}={\left(\frac{p}{q}\right)}^x \\ \Rightarrow \frac{{\left(qp+1\right)}^{p-q}}{{\left(q\right)}^{p-q}}\times \frac{{\left(pq-1\right)}^{p+q}}{{\left(q\right)}^{p+q}}\times \frac{{\left(p\right)}^{p-q}}{{\left(pq+1\right)}^{p-q}}\times \frac{{\left(p\right)}^{p+q}}{{\left(pq-1\right)}^{p+q}}={\left(\frac{p}{q}\right)}^x \\ \Rightarrow \frac{{\left(p\right)}^{p-q}}{{\left(q\right)}^{p-q}}\times \frac{{\left(p\right)}^{p+q}}{{\left(q\right)}^{p+q}}={\left(\frac{p}{q}\right)}^x \\ \Rightarrow \frac{{\left(p\right)}^{p-q+p+q}}{{\left(q\right)}^{p-q+p+q}}={\left(\frac{p}{q}\right)}^x \\ \Rightarrow \frac{{\left(p\right)}^{2p}}{{\left(q\right)}^{2p}}={\left(\frac{p}{q}\right)}^x \\ \Rightarrow {\left(\frac{p}{q}\right)}^{2p}={\left(\frac{p}{q}\right)}^x \\ \Rightarrow 2p=x \\ \Rightarrow x=2p \end{gathered}$$

Hence, if $\frac{{\left(p+{\displaystyle \frac{1}{q}}\right)}^{p-q}{\left(p-{\displaystyle \frac{1}{q}}\right)}^{p+q}}{{\left(q+{\displaystyle \frac{1}{p}}\right)}^{p-q}{\left(q-{\displaystyle \frac{1}{p}}\right)}^{p+q}}={\left(\frac{p}{q}\right)}^x,\mathrm{then}x=\underline{2p}.$