Question

Find the value

(i) ($-7{)}^{12}÷(-7{)}^{12}$ (ii) $7^5$ $÷7^3$ (iii) ${\left(\frac{4}{5}\right)}^3÷{\left(\frac{4}{5}\right)}^2$ (iv) $4^7÷4^5$

Answer 1

It is known that, a^m ÷ aⁿ = a^m−n, where m and n are integers and a is a non-zero rational number.
$$\begin{gathered} \left(\mathrm{i}\right){\left(-7\right)}^{12}÷{\left(-7\right)}^{12} \\ ={\left(-7\right)}^{12-12} \\ ={\left(-7\right)}^0 \\ =1\left(\because a^0=1\right) \end{gathered}$$
$$\begin{gathered} \left(\mathrm{ii}\right)7^5÷7^3 \\ =7^{5-3} \\ =7^2 \\ =7\times 7 \\ =49 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iii}\right){\left(\frac{4}{5}\right)}^3÷{\left(\frac{4}{5}\right)}^2 \\ ={\left(\frac{4}{5}\right)}^{3-2} \\ ={\left(\frac{4}{5}\right)}^1 \\ =\frac{4}{5} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iv}\right)4^7÷4^5 \\ =4^{7-5} \\ =4^2 \\ =4\times 4 \\ =16 \end{gathered}$$