How to simplify the fraction $\frac{(9y^{-2/5})^{3/2}}{(3y^{-1/5})^2}$

Question

I'm stuck on this. I have the answer, just can't work out how to get there. Can someone explain please? $$\large \frac{(9y^{-\frac{2}{5}})^{\frac{3}{2}}}{(3y^{-\frac{1}{5}})^2}$$

Please help, I can not get to the answer, the best I get to is

$$\large \frac{27y^{-\frac{3}{5}}}{9y^{\frac{1}{5}}}$$

Answer 1

I'm guessing you're looking to simplify the expression.

You have a bit of a feel for it, but it's not quite right.

The numerator is correct so far: $27y^{-3/5}$.

In the denominator you need to take care of the exponent:

$$(3y^{-1/5})^2 = 3^2(y^{-1/5})^2 = 9y^{-2/5}.$$

So the expression is so far

$$\frac{27y^{-3/5}}{9y^{-2/5}}$$

Next, use the fact that $y^a/y^b = y^{a-b}$, and you should have a simpler expression.

Answer 2

you can use these $$(kx^a)^b=k^bx^{ab}$$ $$\frac{a^b}{a^c}=a^{b-c}$$

Answer 3

Hint: $\frac{a*b}{c*d}=\frac a c *\frac b d$

Answer 4

Your denominator is wrong because: $$ (3y^{-\frac{1}{5}})^2=9y^{\frac{-2}{5}} $$

so you have:

$$\large \frac{(9y^{-\frac{2}{5}})^{\frac{3}{2}}}{(3y^{-\frac{1}{5}})^2} =\frac{27}{9}y^{-\frac{3}{5}} \cdot y^{\frac{2}{5}} $$

can you do from this?

Answer 5

Okay, here we go: $$\dfrac{(9y^{-\frac{2}{5}})^{\frac{3}{2}}}{(3y^{-\frac{1}{5}})^2}=\dfrac{27y^{-\frac{3}{5}}}{9y^{-\frac{2}{5}}}$$ Divide both sides by $9$ and put the $3$ out front: $$3\dfrac{y^{-\frac{3}{5}}}{y^{-\frac{2}{5}}}$$ remember that $\dfrac{y^a}{y^b}=y^{a-b}$ in order to finally obtain: $$\dfrac{(9y^{-\frac{2}{5}})^{\frac{3}{2}}}{(3y^{-\frac{1}{5}})^2}=3\dfrac{y^{-\frac{3}{5}}}{y^{-\frac{2}{5}}}=3y^{-\frac{3}{5} + \frac{2}{5}}=3y^{\frac{1}{5}}$$