Question

$$\begin{gathered} 5)5^m÷5^{-3}=5^5 \\ 6ii){\left(\frac{5}{8}\right)}^{-7}\times {\left(\frac{8}{5}\right)}^{-4} \end{gathered}$$

Answer 1

Dear student,
Here are the solutions of your asked query:
$$\begin{gathered} \mathrm{Q}.5.5^{\mathrm{m}}÷5^{-3}=5^5 \\ \Rightarrow \frac{5^{\mathrm{m}}}{5^{-3}}=5^5 \\ \Rightarrow 5^{\mathrm{m}-\left(-3\right)}=5^5\{\mathrm{using}\mathrm{the}\mathrm{rule}\frac{{\mathrm{a}}^{\mathrm{m}}}{{\mathrm{a}}^{\mathrm{n}}}={\mathrm{a}}^{\mathrm{m}-\mathrm{n}}\} \\ \Rightarrow 5^{\mathrm{m}+3}=5^5 \\ \Rightarrow \mathrm{m}+3=5\{\mathrm{since}\mathrm{bases}\mathrm{are}\mathrm{same},\mathrm{so}\mathrm{comparing}\mathrm{the}\mathrm{powers}\mathrm{of}\mathrm{both}\mathrm{sides}\} \\ \Rightarrow \mathrm{m}=5-3=2 \\ \mathrm{Q}.6.\left(\mathrm{ii}\right){\left(\frac{5}{8}\right)}^{-7}\times {\left(\frac{8}{5}\right)}^{-4} \\ ={\left(\frac{8}{5}\right)}^7\times {\left(\frac{8}{5}\right)}^{-4}\{\mathrm{using}\mathrm{the}\mathrm{rule}{\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{-\mathrm{m}}={\left(\frac{\mathrm{b}}{\mathrm{a}}\right)}^{\mathrm{m}}\} \\ ={\left(\frac{8}{5}\right)}^{7+\left(-4\right)}\{\mathrm{using}{\mathrm{a}}^{\mathrm{m}}\times {\mathrm{a}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}+\mathrm{n}}\} \\ ={\left(\frac{8}{5}\right)}^3 \\ =\frac{512}{125} \end{gathered}$$
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