Answer
$\displaystyle \frac{3}{4}\cdot x^{1/2}-\frac{5}{3}\cdot x^{-1/2}+\frac{4}{3}\cdot x^{-3/2}$
Work Step by Step
Exponent form:
An expression is in exponent form if
* there are no radicals and
* all powers of unknowns occur in the numerator.
All terms in a sum or difference are of the form:
(constant)(expression with x$)^{p}$
-----------------
First term has a radical, use $a^{m/n}=(a^{1/n})^{m}=\sqrt[n]{a^{m}}$
$\displaystyle \frac{3\sqrt{x}}{4}=\frac{3}{4}\cdot x^{1/2}$
Second term has a radical, use $a^{m/n}=(a^{1/n})^{m}=\sqrt[n]{a^{m}}$
$\displaystyle \frac{5}{3\sqrt{x}}=\frac{5}{3x^{1/2}}\qquad $
...move $x^{1/2}$ out of the denominator,
...using $\displaystyle \quad a^{-n}=\frac{1}{a^{n}}=(\frac{1}{a})^{n}$
$=\displaystyle \frac{5}{3}\cdot x^{-1/2}$
Third term has a radical, use $a^{m/n}=(a^{1/n})^{m}=\sqrt[n]{a^{m}}$
$\displaystyle \frac{4}{3x\cdot x^{1/2}}=\qquad$ ....... use $a^{m}\cdot a^{n}=a^{m+n}$
$\displaystyle \frac{4}{3x^{1+1/2}}=\frac{4}{3}\cdot\frac{1}{x^{3/2}}\qquad $
...move $x^{3/2}$ out of the denominator,
...using $\displaystyle \quad a^{-n}=\frac{1}{a^{n}}=(\frac{1}{a})^{n}$
$=\displaystyle \frac{4}{3}\cdot x^{-3/2}$
The expression, in exponent form is
$\displaystyle \frac{3}{4}\cdot x^{1/2}-\frac{5}{3}\cdot x^{-1/2}+\frac{4}{3}\cdot x^{-3/2}$
