To understand addition of exponents, let's start from a simple example.

Example. Suppose, we want to find ${{2}}^{{3}}\cdot{{2}}^{{4}}$.

We already learned about positive integer exponets, so we can write, that ${{2}}^{{3}}={2}\cdot{2}\cdot{2}$ and ${{2}}^{{4}}={2}\cdot{2}\cdot{2}\cdot{2}$.

So, ${\color{red}{{{{2}}^{{3}}}}}\cdot{\color{green}{{{{2}}^{{4}}}}}={\color{red}{{{2}\cdot{2}\cdot{2}}}}\cdot{\color{green}{{{2}\cdot{2}\cdot{2}\cdot{2}}}}={{2}}^{{7}}$.

Let's see what have we done. We counted number of 2's in ${{2}}^{{3}}$, then counted number of 2's in ${{2}}^{{4}}$. Since we multiplied, then we added number of 2's. Note, that ${3}+{4}={7}$.

It appears, that this rule works not only for positive integer exponents, it works for any exponent.

Rule for adding exponents: $\color{purple}{a^m\cdot a^n=a^{m+n}}$.

Word of caution. It doesn't work, when bases are not equal.

For example, ${{3}}^{{2}}\cdot{{4}}^{{5}}={3}\cdot{3}\cdot{5}\cdot{5}\cdot{5}\cdot{5}\cdot{5}$ which is neither ${{3}}^{{7}}$ nor ${{4}}^{{5}}$.

Word of caution. Above rule doesn't work for addition and subtraction.

For example, ${{2}}^{{3}}+{{2}}^{{4}}\ne{{2}}^{{7}}$, because ${{2}}^{{3}}+{{2}}^{{4}}={8}+{16}={24}$ and ${{2}}^{{7}}={128}$. Clearly, ${24}\ne{128}$.

Example 2. Find ${{2}}^{{3}}\cdot{{2}}^{{-{5}}}$.

It doesn't matter, that exponent is negative.

Just proceed as always: ${{2}}^{{3}}\cdot{{2}}^{{-{5}}}={{2}}^{{{3}+{\left(-{5}\right)}}}={{2}}^{{-{2}}}=\frac{{1}}{{{2}}^{{2}}}=\frac{{1}}{{4}}$.

Even when exponents are fractional, we use the same rule!

Example 3. Find ${{3}}^{{\frac{{1}}{{4}}}}\cdot{{3}}^{{\frac{{2}}{{3}}}}$.

${{3}}^{{\frac{{1}}{{4}}}}\cdot{{3}}^{{\frac{{2}}{{3}}}}={{3}}^{{\frac{{1}}{{4}}+\frac{{2}}{{3}}}}={{3}}^{{\frac{{11}}{{12}}}}={\sqrt[{{12}}]{{{{3}}^{{11}}}}}$.

We can handle radicals, also, because radicals can be rewritten with the help of exponent.

Example 4. Rewrite, using positive exponent: ${\sqrt[{{8}}]{{{3}}}}\cdot{\sqrt[{{7}}]{{\frac{{1}}{{{3}}^{{2}}}}}}$.

First we rewrite numbers, using exponents and then apply the rule:

${\sqrt[{{8}}]{{{3}}}}\cdot{\sqrt[{{7}}]{{\frac{{1}}{{{3}}^{{2}}}}}}={{3}}^{{\frac{{1}}{{8}}}}\cdot{\sqrt[{{7}}]{{{{3}}^{{-{2}}}}}}={{3}}^{{\frac{{1}}{{8}}}}\cdot{{3}}^{{-\frac{{2}}{{7}}}}={{3}}^{{\frac{{1}}{{8}}+{\left(-\frac{{2}}{{7}}\right)}}}={{3}}^{{-\frac{{9}}{{56}}}}=\frac{{1}}{{{{3}}^{{\frac{{9}}{{56}}}}}}$.

Now, it is time to exercise.

Exercise 1. Find ${{3}}^{{2}}\cdot{{3}}^{{5}}$.

Answer: ${{3}}^{{7}}={2187}$.

Exercise 2. Can we use rule for adding exponents to find ${{2}}^{{5}}\cdot{{3}}^{{5}}$?

Answer: No, bases are not equal.

Exercise 3. Find ${{4}}^{{\frac{{1}}{{3}}}}\cdot{{4}}^{{\frac{{2}}{{3}}}}$.

Answer: ${4}$.

Exercise 4. Find ${{3}}^{{2}}\cdot{{3}}^{{-\frac{{1}}{{5}}}}$.

Answer: ${{3}}^{{\frac{{9}}{{5}}}}$.

Exercise 5. Find ${\sqrt[{{7}}]{{\frac{{1}}{{27}}}}}\cdot{\sqrt[{{8}}]{{{9}}}}$.

Answer: ${\sqrt[{{7}}]{{{{3}}^{{-{3}}}}}}\cdot{\sqrt[{{8}}]{{{{3}}^{{2}}}}}=\frac{{1}}{{{3}}^{{\frac{{5}}{{28}}}}}$.