Question

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

Answer 1

Given : $\frac{{(p+\frac{1}{q})}^{p-q}{(p-\frac{1}{q})}^{p+q}}{{(q+\frac{1}{p})}^{p-q}{(q-\frac{1}{p})}^{p+q}}={\left(\frac{p}{q}\right)}^x$

$\Rightarrow \frac{\frac{(pq+1{)}^{p-q}(pq-1{)}^{p+q}}{q^{(p-q)}\times q^{(p+q)}}}{\frac{(pq+1{)}^{p-q}(pq-1{)}^{p+q}}{p^{(p-q)}\times p^{(p+q)}}}={\left(\frac{p}{q}\right)}^x$

$\Rightarrow \frac{(pq+1{)}^{p-q}(pq-1{)}^{p+q}{p}^{(p-q+p+q)}}{(pq+1{)}^{p-q}(pq-1{)}^{p+q}{q}^{(p-q+p+q)}}={\left(\frac{p}{q}\right)}^x$

$\Rightarrow \frac{p^{2p}}{q^{2p}}={\left(\frac{p}{q}\right)}^x$
$\Rightarrow {\left(\frac{p}{q}\right)}^{2p}={\left(\frac{p}{q}\right)}^x$

$\Rightarrow x=2p$
Hence, the required result is $2p$.