### Theory:

In the previous grade, we have learnt about exponent. Let us recall them.

We can write the number $$729$$ as $$9 \times 9 \times 9 = 9^3$$. Here, the number $$9$$ is the base and $$3$$ is the exponent. The exponent is also called as index.

Here, we have found the value of $$9^3$$. Similarly, we can find the value of $$9^{-3}$$ which is the multiplicative inverse of $$9^3$$. That is, $$9^3 \times 9^{-3} = 9^{3 - 3} = 9^0 = 1$$.

Hence, we can write $$9^{-3}$$ as $$9^{-3} = \frac{1}{9^{3}}$$
In general, we can say that $$x^{-n} = \frac{1}{x^n}$$
Consider the numbers with some powers like ${a}^{m},{b}^{n},...$

Here $$a$$ and $$b$$ the base and $$m$$ and $$n$$ are its respective exponents.

The law of exponents are as follows:

$\begin{array}{l}\left(i\right)\phantom{\rule{0.147em}{0ex}}{a}^{m}\cdot {a}^{n}={a}^{m+n}\\ \left(\mathit{ii}\right)\phantom{\rule{0.147em}{0ex}}{\left({a}^{m}\right)}^{n\phantom{\rule{0.147em}{0ex}}}=\phantom{\rule{0.147em}{0ex}}{a}^{\mathit{mn}}\\ \left(\mathit{iii}\right)\phantom{\rule{0.147em}{0ex}}\frac{{a}^{m}}{{a}^{n}}={\left(a\right)}^{m-n},\phantom{\rule{0.147em}{0ex}}\mathit{where}\phantom{\rule{0.147em}{0ex}}m\phantom{\rule{0.147em}{0ex}}>\phantom{\rule{0.147em}{0ex}}n.\\ \left(\mathit{iv}\right)\phantom{\rule{0.147em}{0ex}}{a}^{m}{b}^{m}=\phantom{\rule{0.147em}{0ex}}{\left(\mathit{ab}\right)}^{m}\end{array}$
Example:
1. Consider ${3}^{5}\cdot {3}^{-6}$,

Here the base(3) is the same but the exponents $$5$$ and $$-6$$ are different.

Comparing the property ${a}^{m}\cdot {a}^{n}={a}^{m+n}$ with the expression,

${3}^{5}\cdot {3}^{-6}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}{3}^{5-6}={3}^{-1}=\frac{1}{3}$.

2. Let us take another expression ${\left({9}^{3}\right)}^{5}$.

Here the base (9) is the same, but the exponents $$3$$ and $$5$$ are different.

Comparing the property ${\left({a}^{m}\right)}^{n\phantom{\rule{0.147em}{0ex}}}=\phantom{\rule{0.147em}{0ex}}{a}^{\mathit{mn}}$ with the expression,

${\left({9}^{3}\right)}^{5}={9}^{3×5}={9}^{15}$.

3. Let the next expression be $\frac{{2}^{7}}{{4}^{3}}$.

Here the base values are $$2$$ and $$4 = 2^2$$, and the exponents are $$7$$ and $$3$$.

$\frac{{2}^{7}}{{4}^{3}}=\frac{{2}^{7}}{{\left({2}^{2}\right)}^{3}}$

Applying the property ${\left({a}^{m}\right)}^{n\phantom{\rule{0.147em}{0ex}}}=\phantom{\rule{0.147em}{0ex}}{a}^{\mathit{mn}}$,

$\frac{{2}^{7}}{{\left({2}^{2}\right)}^{3}}=\frac{{2}^{7}}{{2}^{6}}$

Now applying $\phantom{\rule{0.147em}{0ex}}\frac{{a}^{m}}{{a}^{n}}={\left(a\right)}^{m-n},\phantom{\rule{0.147em}{0ex}}\mathit{where}\phantom{\rule{0.147em}{0ex}}m\phantom{\rule{0.147em}{0ex}}>\phantom{\rule{0.147em}{0ex}}n.$

$\frac{{2}^{7}}{{2}^{6}}={2}^{7-6}={2}^{1}=2$