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fracional exponents

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by resilient » Thu Mar 06, 2008 7:14 pm
is the statement 0<u<1 true?

1.0<(u^1/3)<1

2. u^4=1/16


qa is A. but i don't know how to test numbers with fractional exponents. statement 2 is 1/2 or -1/2 but whats up with statement 1.?
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Source: — Data Sufficiency |
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by ikant » Thu Mar 06, 2008 7:57 pm
Hello,

You have already got the stsmt II in control. Let us analyse stsmt I and solve the doubt.

Split the inequality in two inequations and see:

0<u^1/3

Cube on both sides:: 0<u.
So this puts u greater than 0

Next inequality says: u^1/3<1

Again cube on both sides:: u<1

This puts u less than 1.

Uniting the two lets you satisfy the quesn.

I hope this helps
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hmm

by resilient » Thu Mar 06, 2008 8:02 pm
I can see the rational to split them up and its quite clever. OK! I lost you on the operations of cubing the split up parts. Can you explain the steps after you splite them up? This is where you lost me!
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by ikant » Thu Mar 06, 2008 8:27 pm
Well .. you need to take the decision about 'u' while the inequations talk about the cubic root of u.

So i cube both sides of the inequation to introduce u. This allows me to draw inference about it.

I hope youv get it...
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Re: hmm

by Stuart@KaplanGMAT » Fri Mar 07, 2008 12:32 pm
Enginpasa1 wrote:I can see the rational to split them up and its quite clever. OK! I lost you on the operations of cubing the split up parts. Can you explain the steps after you splite them up? This is where you lost me!
fractional exponents are roots.

For example:

4^(1/2) = sqrt(4) = 2

27^(1/3) = cubert(27) = 3

So, let's look at our inequality:

0 < u^(1/3) < 1

To get rid of the exponent (1/3), we can raise every part of the inequality to the exponent (3). That gives us:

0^3 < (u^(1/3))^3 < 1^3

Now, when we raise an exponent to another exponent, we multiple them. So, now we have:

0 < u^(1/3 * 3) < 1

and when we multiply we get:

0 < u^1 < 1

or simply

0 < u < 1.
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