Question

If, $x={\left(\frac{-4}{5}\right)}^{30}÷{\left(\frac{-4}{5}\right)}^{28}$ and $y={\left(\frac{-4}{5}\right)}^2\times {\left(\frac{-4}{5}\right)}^3$, find the value of $(x^3÷y{)}^2$

• A. $\frac{-16}{25}$
• B. $\frac{6}{25}$
• C. $\frac{16}{25}$
• D. $\frac{6}{16}$

Answer 1

The correct option is C

$\frac{16}{25}$

We have:
$x={\left(\frac{-4}{5}\right)}^{30}÷{\left(\frac{-4}{5}\right)}^{28}$
$x={\left(\frac{-4}{5}\right)}^{(30-28)}\because a^m÷a^n=a^{(m-n)}$
$\therefore$ $x={\left(\frac{-4}{5}\right)}^2$
And,
$y={\left(\frac{-4}{5}\right)}^2\times {\left(\frac{-4}{5}\right)}^3$
$y={\left(\frac{-4}{5}\right)}^{(2+3)}\because a^m\times a^n=a^{(m+n)}$
$\therefore$ $y={\left(\frac{-4}{5}\right)}^5$

Now,
$x={\left(\frac{-4}{5}\right)}^2,y={\left(\frac{-4}{5}\right)}^5$
$x^3={\left(\frac{-4}{5}\right)}^{(2\times 3)}\because {\left(a^m\right)}^n=a^{(m\times n)}$
$x^3=(\frac{-4}{5}{)}^6$

So, $x^3÷y=(\frac{-4}{5}{)}^6÷(\frac{-4}{5}{)}^5\Rightarrow (\frac{-4}{5}{)}^{(6-5)}=\frac{-4}{5}\because a^m÷a^n=a^{(m-n)}$

$\therefore (x^3÷y{)}^2=(\frac{-4}{5}{)}^2=\frac{16}{25}$