# Multiplying Reciprocal Exponents

This is the problem I have:

$$y^{\frac{3}{2}} = 5y$$

What I tried so far was raising $y^{\frac{3}{2}}$ to the $\frac{3}{2}$, making it equal $1$, but I had trouble raising $5y$ to that power.

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You have the right idea, you are trying to get rid of the fractional power. However, the best way to do this is to raise each side to the power $2$, so we have:

$$\left(y^{\frac{3}{2}}\right)^{2}=\left(5y\right)^{2}\implies y^{3}=25y^{2},$$

Which is then much easier to manipulate.

Hope this helps!

If you are trying to find all solutions to the equation, then you must solve a cubic equation:

$$y^{3}-25y^{2}=0$$

This can be factorized as:

$$y^{2}\left(y-25\right)=0,$$

Which gives us the following solutions:

$$y=\{0,25\}$$

Where $0$ is a duplicated root.

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Ah, yes. But how would I move on from here? 25 isn't a cube. – Andre Oseguera Aug 21 '12 at 14:22
Of course this does help him, Shaktal. – Babak S. Aug 21 '12 at 14:22
@AndreOseguera Are you trying to find all solutions to the equation? – Shaktal Aug 21 '12 at 14:22

I probably would first write the equation $y^{3/2}=5y$ as $y(y^{1/2})=5y$. This has the solution $y=0$. If $y \ne 0$, we can cancel $y$ from both sides (that is, divide both sides by $y$) and obtain $y^{1/2}=5$. Now square both sides. We get $y=25$.

So the two solutions of the equation are $y=0$ and $y=25$.

Remark: Let us explore your idea. You wanted to get a $y$ on the left-hand side by raising $y^{3/2}$ to a certain power. That can be done, but the appropriate power is $2/3$.

We have $(y^{3/2})^{2/3}=y^{(3/2)(2/3)}=y^1=y$. We need to do the same thing to the right-hand side, and we obtain the equation $y=5^{2/3}y^{2/3}$. Improved the left-hand side, but the right-hand side has been uglified.

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