Question

If $\frac{\mathrm{p}}{\mathrm{q}}={\left(\frac{3}{4}\right)}^{18}÷{\left(\frac{3}{4}\right)}^{16}$, find the value of ${\left(\frac{\mathrm{p}}{\mathrm{q}}\right)}^2$.

Answer 1

Given; $\frac{\mathrm{p}}{\mathrm{q}}={\left(\frac{3}{4}\right)}^{18}÷{\left(\frac{3}{4}\right)}^{16}$.

Now,

$\frac{\mathrm{p}}{\mathrm{q}}={\left(\frac{3}{4}\right)}^{18}÷{\left(\frac{3}{4}\right)}^{16}$

$$\begin{gathered} \Rightarrow \frac{\mathrm{p}}{\mathrm{q}}={\left(\frac{3}{4}\right)}^{18-16}\left[\because {\mathrm{a}}^{\mathrm{m}}÷{\mathrm{a}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}-\mathrm{n}}\right] \\ \Rightarrow \frac{\mathrm{p}}{\mathrm{q}}={\left(\frac{3}{4}\right)}^2 \\ \Rightarrow \frac{\mathrm{p}}{\mathrm{q}}={\left(\frac{3}{4}\right)}^2 \\ \Rightarrow \frac{\mathrm{p}}{\mathrm{q}}=\frac{3}{4}\times \frac{3}{4} \\ \Rightarrow \frac{\mathrm{p}}{\mathrm{q}}=\frac{9}{16} \\ \Rightarrow {\left(\frac{\mathrm{p}}{\mathrm{q}}\right)}^2=\frac{9}{16}\times \frac{9}{16} \\ \Rightarrow {\left(\frac{\mathrm{p}}{\mathrm{q}}\right)}^2=\frac{81}{256} \end{gathered}$$

Hence, ${\left(\frac{\mathrm{p}}{\mathrm{q}}\right)}^2=\frac{81}{256}$.