Question

Simplify the following:

$\left(i\right)3(a^4{b}^3{)}^{10}\times 5(a^2{b}^2{)}^3\left(ii\right)\left(2x^{-2}{y}^3{)}^3\right(iii\left)\frac{(4\times {10}^7)(6\times {10}^{-5})}{8\times {10}^4}\right(iv\left)\frac{4a{b}^2(-5a{b}^3)}{10a^2{b}^2}\right(v\left){\left(\frac{x^2{y}^2}{a^2{b}^3}\right)}^n\right(vi)\frac{(a^{3n-9}{)}^6}{a^{2n-4}}$

Answer 1

(i)$3(a^4{b}^3{)}^{10}\times 5(a^2{b}^2{)}^3$

= $3a^{4\times 10}{b}^{3\times 10}\times 5a^{2\times 3}\times b^{2\times 3}\left\{\begin{array}{cc}\because (a^m{)}^n=a^{mn}& \\ a^m\times a^n=a^{m+n}& \end{array}\right\}$

=$3a^{40}{b}^{30}\times 5a^6{b}^6$

=$3\times 5\times a^{40+6}\times b^{30+6}$

=$15a^{46}{b}^{36}$

(ii) $(2x^{-2}{y}^3{)}^3=2^3\times x^{-2\times 3}\times y^{3\times 3}$ $\left\{\begin{array}{cc}\because (a^m{)}^n=a^{mn}& \\ a^m\times a^n=a^{m+n}& \end{array}\right\}$

=$8\times x^{-6}\times y^9=8x^{-6}{y}^9$

(iii) $\frac{(4\times {10}^7)(6\times {10}^{-5})}{8\times {10}^4}=\frac{4\times 6}{8}{10}^{7-5-4}$ $\left\{\begin{array}{cc}\because a^m\times a^n=a^{m+n}& \\ a^{\circ }=1& \end{array}\right\}$

$=3\times {10}^{-2}=\frac{3}{{10}^2}=\frac{3}{10\times 10}=\frac{3}{100}$

(iv) $\frac{4a{b}^2(-5a{b}^3)}{10a^2{b}^2}=\frac{4\times (-5)}{10}\times a^{1+1-2}{b}^{2+3-2}$ $\left\{\begin{array}{cc}\because a^m\times a^n=a^{m+n}& \\ a^{\circ }=1& \end{array}\right\}$

$=-2\times a^0{b}^3=-2\times 1\times b^3=-2b^3$

(v) ${\left(\frac{x^2{y}^2}{a^2{b}^3}\right)}^n=\frac{x^{2n}\times y^{2n}}{a^{2n}{b}^{3n}}=\frac{x^{2n}.y^{2n}}{a^{2n}{b}^{3n}}\left\{\begin{array}{c}\because (a^m{)}^n=a^{mn}\end{array}\right\}$

(vi) $\frac{(a^{3n-9}{)}^6}{a^{2n-4}}\left\{\begin{array}{cc}\because a^m\times a^n=a^{m+n}& \\ (a^m{)}^n=a^{mn}& \end{array}\right\}=\frac{a^{18n-54}}{a^{2n-4}}=a^{18n-54-2n+4}=a^{16n-50}$