How do you evaluate ${{5}^{3}}{{.5}^{2}}$ ?

Hint: We are given term as \[{{5}^{3}}\] and ${{5}^{2}}$ , we are asked to find the product of three two, to do so we will learn about the type of expression we have. We will then learn about the exponential term, we will be using ${{x}^{a}}.{{x}^{b}}={{x}^{a+b}}$ . And we will also find the solution by just spending the term and then finding the product. We will also learn about various properties of the exponential function which are useful in solving such problems.

Complete step by step answer:
We are given two terms as ${{5}^{3}}$ and ${{5}^{2}}$ . We can see that in their terms, we have a number which is reused to some term. The term where one number is raised to some power like ${{a}^{b}}$ is called exponential form, so we are given two terms into exponential form.
We have to evaluate the product of ${{5}^{3}}$ and ${{5}^{2}}$ to find the product. We will expand the terms and we will find the product.
Now as we have ${{5}^{3}}$ , it has power 3, so it means 5 is multiplied by itself 3 times. So,
${{5}^{3}}=5\times 5\times 5$
Which is the same as 125.
 $\Rightarrow {{5}^{3}}=125$ ………………………… (1)
Now in ${{5}^{2}}$ , it has power 2, so it means 5 is multiplied by itself two times. So,
$\Rightarrow {{5}^{2}}=5\times 5$……………………. (2)
Now using equation (1) and (2) we get –
$\Rightarrow {{5}^{3}}{{.5}^{2}}=125\times 25$
 Now we multiply 125 and 25 we get –
$\Rightarrow {{5}^{3}}{{.5}^{2}}=3125$
Hence, we get ${{5}^{3}}{{.5}^{2}}=3125$ .
If the exponent has a different base then we do as we discussed above. If the exponents have the same base then its simplification will be a little easier.
We have rules for the simplification of such an exponent.
In our problem, we have ${{5}^{3}}{{.5}^{2}}$ , they have the same base.
So, we can use the identity ${{x}^{a}}\times {{x}^{b}}={{x}^{a+b}}$ .
So, $\Rightarrow {{5}^{3}}\times {{5}^{2}}={{5}^{3+2}}$
Simplifying, we get –
$={{5}^{5}}$ .
As ${{5}^{5}}$ is 3125.
So, ${{5}^{3}}{{.5}^{2}}=3125$

Note: There are other identities which are working for the same base or for the general exponential form of the number.
We have ${{x}^{a}}\times {{x}^{b}}={{x}^{a+b}}$
In case of division $\dfrac{{{x}^{a}}}{{{x}^{b}}}={{x}^{a-b}}$
In case of inverse $\dfrac{1}{{{x}^{a}}}={{x}^{-a}}$
In case of power raise to whole ${{\left( {{x}^{a}} \right)}^{b}}={{x}^{a\times b}}$ .