Chapter 7.2, Problem 71PFA

To determine

To simplify: the given all expressions.

Answer to Problem 71PFA

  $\frac{3p^{-5}{q}^4{r}^7}{9p^8{q}^{-9}{r}^{-3}}$

Explanation of Solution

Given information:

  $\begin{array}{l}\left(A\right)\frac{4p^5{q}^{-4}{r}^{-7}}{12p^{-8}{q}^9{r}^3}\\ \left(B\right)\frac{2p^{-5}{q}^4{r}^{-7}}{6p^8{q}^{-9}{r}^3}\\ \left(C\right)\frac{3p^{-5}{q}^4{r}^7}{9p^8{q}^{-9}{r}^{-3}}\\ \left(D\right)\frac{5p^5{q}^{-4}{r}^7}{15p^{-8}{q}^9{r}^{-3}}\end{array}$

Concept used:

  $\begin{array}{l}{a}^{-m}=\frac{1}{a^m}\\ a^m\times a^n=a^{m+n}\\ \frac{a^m}{a^n}=a^{m-n}\end{array}$

Calculation:

On solving the option (A), it gives

  $\begin{array}{l}\left(A\right)\frac{4p^5{q}^{-4}{r}^{-7}}{12p^{-8}{q}^9{r}^3}\\ \Rightarrow \frac{p^5\times p^8}{3q^9\times q^4\times r^3\times r^7}\mathrm{.....}\left[\because a^{-m}=\frac{1}{a^m}\right]\\ \Rightarrow \frac{p^{5+8}}{3q^{9+4}\times r^{3+7}}\mathrm{.......}\left[\because a^m\times a^n=a^{m+n}\right]\\ \Rightarrow \frac{p^{13}}{3q^{13}{r}^{10}}\end{array}$

On solving the option (B), it gives

  $\begin{array}{l}\left(B\right)\frac{2p^{-5}{q}^4{r}^{-7}}{6p^8{q}^{-9}{r}^3}\\ \Rightarrow \frac{q^4\times q^9}{3p^8\times p^5\times r^3\times r^7}\mathrm{.....}\left[\because a^{-m}=\frac{1}{a^m}\right]\\ \Rightarrow \frac{q^{4+9}}{3p^{8+5}\times r^{3+7}}\mathrm{.......}\left[\because a^m\times a^n=a^{m+n}\right]\\ \Rightarrow \frac{q^{13}}{3p^{13}{r}^{10}}\end{array}$

On solving the option (C), it gives

  $\begin{array}{l}\left(C\right)\frac{3p^{-5}{q}^4{r}^7}{9p^8{q}^{-9}{r}^{-3}}\\ \Rightarrow \frac{q^4\times q^9\times r^7\times r^3}{3p^8\times p^5}\mathrm{.....}\left[\because a^{-m}=\frac{1}{a^m}\right]\\ \Rightarrow \frac{q^{4+9}\times r^{7+3}}{3p^{8+5}}\mathrm{.......}\left[\because a^m\times a^n=a^{m+n}\right]\\ \Rightarrow \frac{q^{13}{r}^{10}}{3p^{13}}\end{array}$

On solving the option (D), it gives

  $\begin{array}{l}\left(D\right)\frac{5p^5{q}^{-4}{r}^7}{15p^{-8}{q}^9{r}^{-3}}\\ \Rightarrow \frac{p^5\times p^8\times r^7\times r^3}{3q^9\times q^4}\mathrm{.....}\left[\because a^{-m}=\frac{1}{a^m}\right]\\ \Rightarrow \frac{p^{5+8}\times r^{7+3}}{3q^{9+4}}\mathrm{.......}\left[\because a^m\times a^n=a^{m+n}\right]\\ \Rightarrow \frac{p^{13}{r}^{10}}{3q^{13}}\end{array}$