How to simplify; $\dfrac{{{t}^{2}}}{{{t}^{-2}}}$?

Hint: In this question we have been given with a term which is in the form of a fraction. The numerator and the denominator of the fraction consists of terms in exponents. We will use the property of exponent which is $\dfrac{1}{{{a}^{-b}}}={{a}^{b}}$ and then get the denominator of the fraction in the form of multiplication, eliminating the denominator. We will then use the property of exponents ${{a}^{b}}\times {{a}^{c}}={{a}^{b+c}}$ on the terms and simplify the expression to get the required solution.

Complete step by step solution:
We have the expression given to us as:
$\Rightarrow \dfrac{{{t}^{2}}}{{{t}^{-2}}}$
We can split the fraction and write it in the form of multiplication as:
$\Rightarrow {{t}^{2}}\times \dfrac{1}{{{t}^{-2}}}$
Now we know the property of exponents that $\dfrac{1}{{{a}^{-b}}}={{a}^{b}}$ therefore on using this property on the fraction part, we get:
$\Rightarrow {{t}^{2}}\times {{t}^{2}}$
Now we can see that there are two terms in multiplication and both the exponential terms have the same base which is $t$ therefore, we can use the property of exponents that ${{a}^{b}}\times {{a}^{c}}={{a}^{b+c}}$ on the terms and write it as:
$\Rightarrow {{t}^{2+2}}$
On simplifying the exponent term, we get:
$\Rightarrow {{t}^{4}}$, which is the required solution.

Note: It is to be remembered that exponential equations are equations that have in them exponential terms present. The properties of exponent such as $\dfrac{1}{{{a}^{-b}}}={{a}^{b}}$ and ${{a}^{b}}\times {{a}^{c}}={{a}^{b+c}}$ are true for all the real numbers. The inverse of the property can be also be used such as ${{a}^{b}}=\dfrac{1}{{{a}^{-b}}}$ and ${{a}^{b+c}}={{a}^{b}}\times {{a}^{c}}$ to split the terms. One of the most commonly used exponential functions is ${{e}^{x}}$which has its base as an irrational number.