Question

Simplify the following:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Answer 1

(i)
$$\begin{gathered} 3{\left(a^4{b}^3\right)}^{10}\times 5{\left(a^2{b}^2\right)}^3 \\ =3\times a^{40}\times b^{30}\times 5\times a^6\times b^6 \\ =15\times a^{40}\times a^6\times b^{30}\times b^6 \\ =15\times a^{40+6}\times b^{30+6}\left[a^m\times a^n=a^{m+n}\right] \\ =15a^{46}{b}^{36} \end{gathered}$$

(ii)
$$\begin{gathered} {\left(2x^{-2}{y}^3\right)}^3 \\ =2^3\times {\left(x^{-2}\right)}^3\times {\left(y^3\right)}^3 \\ =8\times x^{-6}\times y^9 \\ =8x^{-6}{y}^9 \end{gathered}$$

(iii)
$$\begin{gathered} \frac{\left(4\times {10}^7\right)\left(6\times {10}^{-5}\right)}{8\times {10}^4} \\ =\frac{4\times {10}^7\times 6\times {10}^{-5}}{8\times {10}^4} \\ =\frac{24\times {10}^{7+\left(-5\right)}}{8\times {10}^4} \\ =\frac{24\times {10}^2}{8\times {10}^4} \end{gathered}$$
$$\begin{gathered} =\frac{24\times {10}^2\times {10}^{-4}}{8} \\ =3\times {10}^{2+\left(-4\right)} \\ =3\times {10}^{-2} \\ =\frac{3}{100} \end{gathered}$$

(iv)
$$\begin{gathered} \frac{4a{b}^2\left(-5a{b}^3\right)}{10a^2{b}^2} \\ =\frac{4\times a\times b^2\times \left(-5\right)\times a\times b^3}{10a^2{b}^2} \\ =\frac{-20\times a^1\times a^1\times b^2\times b^3}{10a^2{b}^2} \end{gathered}$$
$$\begin{gathered} =\frac{-20\times a^{1+1}\times b^{2+3}}{10a^2{b}^2} \\ =-2\times a^2\times b^5\times a^{-2}\times b^{-2} \\ =-2\times a^{2+\left(-2\right)}\times b^{5+\left(-2\right)} \\ =-2\times a^0\times b^3 \\ =-2b^3 \end{gathered}$$

(v)
$$\begin{gathered} {\left(\frac{x^2{y}^2}{a^2{b}^3}\right)}^n \\ =\frac{{\left(x^2\right)}^n{\left(y^2\right)}^n}{{\left(a^2\right)}^n{\left(b^3\right)}^n} \\ =\frac{x^{2n}{y}^{2n}}{a^{2n}{b}^{3n}} \end{gathered}$$

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