Question

Simplify, giving solution with positive index​ :(xxii) $\left(7p^2{q}^9{r}^5{)}^2\right(4pqr{)}^3÷(14p^6{q}^{10}{r}^4{)}^2$

Answer 1

Given : $\left(7p^2{q}^9{r}^5{)}^2\right(4pqr{)}^3÷(14p^6{q}^{10}{r}^4{)}^2$

It can be written as:

$\frac{\left(7p^2{q}^9{r}^5{)}^2\right(4pqr{)}^3}{(14p^6{q}^{10}{r}^4{)}^2}=\frac{\left(7p^2{q}^{9\times 2}{r}^{5\times 2}\right)\left(4^3{p}^3{q}^3{r}^3\right)}{{14}^2{p}^{6\times 2}{q}^{10\times 2}{r}^{4\times 2}}$

[$\because$ $(a^m{)}^n=a^{mn}$]
On further calculation , we have :

$\frac{\left(7p^2{q}^9{r}^5{)}^2\right(4pqr{)}^3}{(14p^6{q}^{10}{r}^4{)}^2}=\frac{7\times 7p^4{q}^{18}{r}^{10}(4\times 4\times 4p^3{q}^3{r}^3)}{2\times 7\times 2\times 7\times p^{12}{q}^{20}{r}^8}$

$=p^{4-12+3}{q}^{18-20+3}{r}^{10-8+3}4\times 4$
$=16p^{-5}q{r}^5=\frac{16q{r}^5}{p^5}$

[$\because$ $a^{-m}=\frac{1}{a^m}$]

Hence, $\frac{\left(7p^2{q}^9{r}^5{)}^2\right(4pqr{)}^3}{(14p^6{q}^{10}{r}^4{)}^2}=\frac{16q{r}^5}{p^5}$