Question

Simplify: [4 MARKS]

i)${\left(\frac{5}{8}\right)}^{-7}\times {\left(\frac{8}{5}\right)}^{-5}$
ii)$\frac{(2^5{)}^2\times 7^3}{8^3\times 7}$

Answer 1

Each: [2 Marks]

i)${\left(\frac{5}{8}\right)}^{-7}\times {\left(\frac{8}{5}\right)}^{-5}$

$=\frac{5^{-7}}{8^{-7}}\times \frac{8^{-5}}{5^{-5}}$

$=\frac{5^{-7}}{5^{-5}}\times \frac{8^{-5}}{8^{-7}}$

$=5^{(-7)-(-5)}\times 8^{(-5)-(-7)}$

($\because \frac{a^m}{a^n}=a^{m-n}$)

$=5^{-2}\times 8^2=\frac{8^2}{5^2}=\frac{64}{25}$

ii)$\frac{(2^5{)}^2\times 7^3}{8^3\times 7}$

$\frac{(2^5{)}^2\times 7^3}{8^3\times 7}$

= $\frac{\left(2^{5\times 2}\right)\times 7^3}{(2^3{)}^3\times 7}$

$(\because (a^m{)}^n=a^{m\times n})$

= $\frac{2^{10}\times 7^3}{2^9\times 7}$

=$2^{10-9}\times 7^{3-1}=2\times 7^2$

=$2\times 49=98$