Question

Let $a={\left(\frac{1}{9}\right)}^{-2{\log }_{3}7}$ and $b=2^{-{\log }_{\frac{1}{2}}(7)}$ then $a=(b{)}^k$ where $k$ is equal to

• A. 4
• B. 3
• C. 2
• D. 1

Answer 1

Answer: (A)

$a=9^{2{\log }_3(7)}=9^{{\log }_3(49)}=3^{2{\log }_3(49)}=3^{{\log }_3(2401)}=2401$
$b=2^{{\log }_2(7)=7}$
$\therefore a=b^4\Rightarrow k=4A$