how so simplify this exponential equations

Question

((a^3/2)/(b^3))/((a^-1)/(b^2))

I tried to solve this problem many times, however I tend to get the wrong answer.

Here is the method I tried

(((a^3)^1/2)/(b^3))*... sorry I get confused

i got

(a^2)/(b)

Answer 1

This is a slower explanation. For a fast solution, check the other answer.

Based on your first simplification ($a^{3/2} = (a^{1/2})^{3}$), I take the equation reads

$$((a^{3/2})/(b^3))/((a^{-1})/(b^2))$$ Lets write this out more nicely with fractions, but a division symbol in the middle:

$$\frac{a^{3/2}}{b^3} \div\frac{a^{-1}}{b^2}$$ Instead of dividing by that fraction on the right, we multiply with its inverse. The above expression then reads $$\frac{a^{3/2}}{b^3} \cdot \frac{b^2}{a^{-1}}$$ We multiply numerator and denominator. $$\frac{a^{3/2}\cdot b^2}{b^3\cdot a^{-1}}$$ We can simplify the $b$ expression:$$\frac{b^2}{b^3} =\frac{\color{red}{b^2}}{\color{red}{b^2}\cdot b} = \frac{1}{b}$$ Or, more generally using the law $$\frac{x^a}{x^b} = x^{a-b}$$ Thus, we can also simplify $$\frac{a^{3/2}}{a^{-1}} = a^{3/2 - (-1)} = a^{3/2 + 1} = a^{3/2 + 2/2} = a^{5/2}$$ And the entire expression becomes $$\frac{\color{blue}{a^{3/2}}\cdot \color{red}{b^2}}{\color{red}{b^3}\cdot \color{blue}{a^{-1}}} = \frac{\color{blue}{a^{5/2}}}{\color{red}{b}}$$ Which is the final expression.

For completness: There are lots of ways to get to that answer, e.g. one could simplify $$\frac{a^{3/2}}{a^{-1}}$$ using the law $$a^{-n} = \frac{1}{a^n}$$ which here implies that $$\frac{1}{a^{-1}} = \frac{1}{\frac{1}{a}} = a$$ And then $$x^a\cdot x^b = x^{a+b}$$ to get $$\frac{a^{3/2}}{a^{-1}} = a^{3/2}\cdot a = a^{5/2}$$

Answer 2

If you have:

$$ \frac{\;\;\frac{a^{3/2}}{b^3}\;\;}{\frac{a^{-1}}{b^2}} $$ then it is: $$ \frac{a^{3/2}}{b^3}\frac{b^2}{a^{-1}}=\frac{a^{3/2}\cdot a}{b}=\frac{a^{5/2}}{b} $$

Answer 3

If you have: $$((a^3/2)/(b^3))/((a^{-1})/(b^2))$$ $$=\dfrac{\dfrac{a^3}{2b^3}}{\dfrac{a^{-1}}{b^2}}$$ $$=\dfrac{a^3b^2}{2a^{-1}b^3}$$ $$=\dfrac{a^3}{2a^{-1}b}$$ $$=\dfrac{a^4}{2b}$$