Question

$\sqrt{m^4{n}^4}\times \sqrt[6]{6m^2{n}^2}\times \sqrt[3]{3m^2{n}^2}={\left(mn\right)}^k$, then find the value of $k$.

• A. $6$
• B. $3$
• C. $2$
• D. $1$

Answer 1

Answer: (B)

$3$

$\sqrt{m^4{n}^4}\times \sqrt[6]{6m^2{n}^2}\times \sqrt[3]{3m^2{n}^2}={\left(mn\right)}^k$

$\Rightarrow {\left(mn\right)}^{4/2}\times {\left(mn\right)}^{2/6}\times {\left(mn\right)}^{2/3}={\left(mn\right)}^k$

$\Rightarrow {\left(mn\right)}^{2+\frac{1}{3}+\frac{2}{3}}={\left(mn\right)}^k$

$\Rightarrow k=2+\frac{1}{3}+\frac{2}{3}$

$\therefore$ $k=3$.