Simplifying with exponents

Question

So somehow I made it all the way to Calc II but struggle when it comes to this basic thing (not exactly the most 100% solid algebra foundation it seems).

Unable to simplify this further:
$$ =\frac{1}{3x^{\frac{2}{3}}}+\frac{2}{3x^{\frac{5}{3}}} $$

to $$ =\frac{x+2}{3x^{\frac{5}{3}}} $$

Really not sure how to pull out the fractions in the exponent to further simplify it. Not exactly the greatest thing to be able to integrate by parts and work on power series but then be unable to simplify this...

Answer 1

Recall that $x^{a}\cdot x^{b}=x^{a+b}$. Thus, to simplify your sum of fractions, we bring both of the original fractions to a common denominator using the aforementioned rule.

Note that $3x^{\frac{2}{3}}\cdot x=3x^{\frac{2}{3}+1}=3x^{\frac{5}{3}}$. Thus,

$$ \frac{1}{3x^{\frac{2}{3}}}+\frac{2}{3x^{\frac{5}{3}}}=\frac{1\cdot x}{3x^{\frac{2}{3}}\cdot x}+\frac{2}{3x^{\frac{5}{3}}}=\frac{x}{3x^{\frac{5}{3}}}+\frac{2}{3x^{\frac{5}{3}}}=\frac{x+2}{3x^{\frac{5}{3}}}. $$

Answer 2

When you multiply terms, you add exponents, so $x \cdot x^{\frac 23}=x^{\frac 53}$ They multiplied top and bottom of the first by $x$ to put the two over a common denominator.

Answer 3

You can treat $\frac{1}{3x^{2/3}}$ as $3x^{-2/3}$

Similarly, you can treat $\frac{2}{3x^{5/3}}$ as $2*3x^{-5/3}=6x^{-5/3}$