# Help with handling fractional powers in equations

1. Sep 4, 2011

### JB34

Hello, I'm pretty rusty when it comes to rearranging more complex equations and can't seem to remember how to deal with fractions as powers, for example;

[1/2] x y^[1/2] = sqrt[[z] over [x y^-1/2]]

and

[2/3] x y^[-1/3] = sqrt[[z] over [x y^2/3]]

I'm trying to solve a similar, but more complex, equation and am really stuck on how the fractions should be best handled in problems of this type. If someone could give me some guidance on solving my examples for x it might make things click for me!

Last edited: Sep 4, 2011
2. Sep 4, 2011

### gsal

I am not sure what your equations are supposed to look like...either my browser is not rendering them well or something, but they don't even make sense to me, can't make out much...are you attempting to use latex or something? can you attempt to express them with simple text and a few parenthesis?

3. Sep 4, 2011

### JB34

Sorry, it's shorthand from an equation editor, quite similar to latex, I'm so used to reading it that I didn't think to make it clearer

So there are three variables here, x, y & z in each equation but the equations do not complement each other, they are individual examples... my first attempt at latex:

$\frac{1}{2}x y^{{1}/{2}} = \sqrt{\frac{z}{{x y^{-1/2}}}}$

and

$\frac{2}{3}x y^{{-1}/{3}} = \sqrt{\frac{z}{{x y^{2/3}}}}$

Last edited: Sep 4, 2011
4. Sep 4, 2011

### HallsofIvy

The first thing I would do is get rid of the square roots by squaring both sides of the equations. Now, the question is, what are you trying to do with the equations? Solve for one of the variables? Which one?

Do you know the "laws of exponents"? $a^m a^n= a^{m+ n}$, $(a^m)^n= a^{mn}$, $a^{-1}= 1/a$. You will need to use those.

5. Sep 4, 2011

### jambaugh

Also note: $x^{a/b} = (x^a)^{1/b} = \sqrt{x^a}$

This format comes from the "power of a power" rule. For x > 0 ....
$(x^{1/n})^n = x^{n \cdot 1/n } = x^1 = x$
and also $(\sqrt[n]{x})^n = x$
so reciprocal powers must be the same as roots.

6. Sep 4, 2011

### JB34

As in my original post I'm looking at solving for x.

Thanks for the pointers so far, I am familiar with the laws of exponents but like I said I'm very rusty, I probably haven't had to solve an equation with fractions for 10 years!

7. Sep 4, 2011

### JB34

Right so taking my first one...

$\frac{1}{2}x y^{{1}/{2}} = \sqrt{\frac{z}{{x y^{-1/2}}}}$

I was thinking along the lines of...

dividing both sides by $y^{{1}/{2}}$ gives:

$\frac{1}{2} x = \sqrt{\frac{z}{{x y}}}$

multiplying both sides by $2 x$ gives:

$x^{2} = \sqrt{\frac{2 z}{y}}$

square root both sides to give:

$x = \sqrt[3]{\frac{2z}{{y}}}$

Have I lost the plot or is that solution sound?

Last edited: Sep 4, 2011
8. Sep 4, 2011

### gsal

I think you have made a mistake in the very first step...

when you divide by y1/2, you have the term on the left side of the equation just fine; but on the right one, you are forgetting to square y1/2 before you bringing into the square root already there, so that you can then add the exponents with the y-1/2 in there...needless to say, it does not work out that way you have it.

9. Sep 4, 2011

### JB34

Looking back at it you're right, I did say I was rusty!

I've only got a couple of days to try and get my head around this ... not looking good :(

$\frac{1}{2}x y^{{1}/{2}} = \sqrt{\frac{z}{{x y^{-1/2}}}}$

ok so back to the beginning...

remove the squareroot by squaring both sides...

$(\frac{1}{2})^{2}x^{2} y = \frac{z}{{x y^{-1/2}}}$

simplified...

$\frac{1}{4}x^{2} y = \frac{z}{{x y^{-1/2}}}$

multiply by 4 x...

$x^{3} y = \frac{4 z}{{y^{-1/2}}}$

divide by $y$...

$x^{3} = \frac{4 z}{{y^{1/2}}}$

cube root...

$x = \sqrt[3]{\frac{4 z}{{y^{1/2}}}}$

Hmmm, any feedback, this still doesn't feel right.

Last edited: Sep 4, 2011
10. Sep 4, 2011

### JB34

If someone could go through one of the two examples, how they would tackle it, that really would help, ultimately I only need to solve a couple of problems and then I probably won't use fractional powers for years again

Last edited: Sep 4, 2011