Question

How do you simplify ${{2}^{\dfrac{5}{2}}}-{{2}^{\dfrac{3}{2}}}$?

Answer 1

Hint: ${{2}^{\dfrac{5}{2}}}-{{2}^{\dfrac{3}{2}}}$ is a difference of two fractional exponents . We can first simplify the fractional exponents and then perform the operation but it will be difficult and time – taking . So, we will use a simpler method to simplify this . In general , ${{b}^{p+q}}={{b}^{p}}\cdot {{b}^{q}}$. So, we will use this to simplify ${{2}^{\dfrac{5}{2}}}$ which can also be written as${{2}^{\dfrac{3}{2}+\dfrac{2}{2}}}$ and hence by using the above equation , we get
${{2}^{\dfrac{5}{2}}}={{2}^{\dfrac{3}{2}+\dfrac{2}{2}}}={{2}^{\dfrac{3}{2}}}\cdot {{2}^{\dfrac{2}{2}}}$ .
We have to use this in the solution.

Complete step by step solution:
We know ,
${{2}^{\dfrac{5}{2}}}={{2}^{\dfrac{3}{2}+\dfrac{2}{2}}}={{2}^{\dfrac{3}{2}}}\cdot {{2}^{\dfrac{2}{2}}}$
Therefore, ${{2}^{\dfrac{5}{2}}}-{{2}^{\dfrac{3}{2}}}$ can also be written as
${{2}^{\dfrac{3}{2}}}\cdot {{2}^{\dfrac{2}{2}}}-{{2}^{\dfrac{3}{2}}}={{2}^{\dfrac{3}{2}}}({{2}^{\dfrac{2}{2}}}-1)$
 $={{2}^{\dfrac{3}{2}}}(2-1)$ , since ${{2}^{\dfrac{2}{2}}}={{2}^{1}}=2$
 $={{2}^{\dfrac{3}{2}}}$
And since ,
${{2}^{\dfrac{3}{2}}}=({{2}^{\dfrac{2}{2}+\dfrac{1}{2}}})$
And hence it can also be written as
${{2}^{\dfrac{3}{2}}}=({{2}^{\dfrac{2}{2}+\dfrac{1}{2}}})={{2}^{\dfrac{2}{2}}}\cdot {{2}^{\dfrac{1}{2}}}=2\cdot {{2}^{\dfrac{1}{2}}}$

Therefore ,
${{2}^{\dfrac{5}{2}}}-{{2}^{\dfrac{3}{2}}}=2\times \sqrt{2}$.

Note: There are many ways to simplify a fractional exponent . The exponent of a number says how many times to use the number in a multiplication. But fractional exponents are those which are the exponents that are in the form of fraction. For example An exponent of \[{\scriptscriptstyle 1\!/\!{ }_2}\]is actually square root , an exponent of \[1/3\] is cube root , an exponent of \[1/4\] is $4^{th}$ root and so on . It is important to have a knowledge of laws of exponents to simplify exponents and fractional exponents.