Question

If $\frac{{\left(a+\frac{1}{b}\right)}^p\times {\left(a-\frac{1}{b}\right)}^q}{{\left(b+\frac{1}{a}\right)}^p\times {\left(b-\frac{1}{a}\right)}^q}={\left(\frac{a}{b}\right)}^x$ then waht is x ?

• A. $p+q$
• B. $p-q$
• C. $pq$
• D. None of these

Answer 1

The correct option is A $p+q$

$\frac{{(a+\frac{1}{b})}^p\times {(a-\frac{1}{b})}^q}{{(b+\frac{1}{a})}^p\times {(b-\frac{1}{9})}^q}$

$\Rightarrow \frac{{\left(\frac{ab+1}{b}\right)}^p{\left(\frac{ab-1}{b}\right)}^q}{{\left(\frac{ab+1}{a}\right)}^p{\left(\frac{ab-1}{a}\right)}^q}$

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

$x=p+q$.