Question

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

• A. $p-q$
• B. $p+q$
• C. $q-p$
• D. $pq$

Answer 1

Answer: (B)

$p+q$

Given,

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

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

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

$\frac{p^{p+q}}{q^{p+q}}={\left(\frac{p}{q}\right)}^x$ ____ $\because a^m.a^n=a^{m+n}$

${\left(\frac{p}{q}\right)}^{p+q}={\left(\frac{p}{q}\right)}^x$ ____ $\because \frac{a^m}{b^m}={\left(\frac{a}{b}\right)}^m$

$\therefore x=p+q$