Question

Simplify.

(i) 7⁴ × 7² (ii) ($-$11)⁵ × ($-$11)² (iii) ${\left(\frac{6}{7}\right)}^3\times {\left(\frac{6}{7}\right)}^5$


(iv) ${\left(-\frac{3}{2}\right)}^5\times {\left(-\frac{3}{2}\right)}^3$ (v) $a^{16}\times a^7$ (vi) ${\left(\frac{\mathrm{P}}{5}\right)}^3\times {\left(\frac{\mathrm{P}}{5}\right)}^7$

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.
$$\begin{gathered} \left(\mathrm{i}\right)7^4\times 7^2 \\ =7^{4+2} \\ =7^6 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{ii}\right){\left(-11\right)}^5\times {\left(-11\right)}^2 \\ ={\left(-11\right)}^{5+2} \\ ={\left(-11\right)}^7 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iii}\right){\left(\frac{6}{7}\right)}^3\times {\left(\frac{6}{7}\right)}^5 \\ ={\left(\frac{6}{7}\right)}^{3+5} \\ ={\left(\frac{6}{7}\right)}^8 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iv}\right){\left(-\frac{3}{2}\right)}^5\times {\left(-\frac{3}{2}\right)}^3 \\ ={\left(-\frac{3}{2}\right)}^{5+3} \\ ={\left(-\frac{3}{2}\right)}^8 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{v}\right)a^{16}\times a^7 \\ =a^{16+7} \\ =a^{23} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{vi}\right){\left(\frac{p}{5}\right)}^3\times {\left(\frac{p}{5}\right)}^7 \\ ={\left(\frac{p}{5}\right)}^{3+7} \\ ={\left(\frac{p}{5}\right)}^{10} \end{gathered}$$