Question

If $x={\left(\frac{3}{2}\right)}^2{\left(\frac{2}{3}\right)}^{-4}$, the value of $x^{-2}$ is?

• A. ${\left(\frac{2}{3}\right)}^{12}$
• B. ${\left(\frac{3}{2}\right)}^{12}$
• C. ${\left(\frac{6}{3}\right)}^{-12}$
• D. ${\left(\frac{3}{6}\right)}^{-12}$

Answer 1

The correct option is A ${\left(\frac{2}{3}\right)}^{12}$

we know,

$\frac{a^{-m}}{b^{-m}}=(\frac{a}{b}{)}^{-m}=(\frac{b}{a}{)}^m$

so,

If $x={\left(\frac{3}{2}\right)}^2{\left(\frac{2}{3}\right)}^{-4}$

$=(\frac{3}{2}{)}^2$ $\ast (\frac{3}{2}{)}^4$

$=(\frac{3}{2}{)}^6$

we know,

$(a^m{)}^n=a^{mn}$

so,

$x^{-2}$

$=(\frac{3}{2}{)}^{-12}$

$=(\frac{2}{3}{)}^{12}$