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Is x^5 > x^4

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Is x^5 > x^4

by faraz_jeddah » Tue Aug 27, 2013 11:59 am
Is x^5 > x^4?


1 - x^3 > -x
2 - 1/x < x

I want to use the algebraic approach here.

Target question can be rephrased as
Is x^4(x-1) > 0?

Which means either
x^4>0 and x>1 (can I summarize this as x>1?)
OR
x^4<0 and x<1 (Can I summarize this as x<0?)

So Now when I try to evaluate the statements, I should try to prove either x>1 or x<0

Statement 1
x^3 > - x
This can be true when x>1 and not true when x is a negative fraction.

Statement 2
1/x < x
This can be true when x>1 and not true when x is a positive fraction.

So when I combine the statements I find the 'common' condition is x>1 which is sufficient.

I want to know if this approach is correct or if I have overlooked anything.

Thanks.
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by ceilidh.erickson » Tue Aug 27, 2013 12:27 pm
I think you've gotten a bit mixed up with your target question here. If we want to simplify this algebraically, we could say:

(x^4)(x^1) > x^4 ?

It's tempting to divide both sides by x^4 here... but what if x = 0 ? We're not allowed to divide by 0. For non-zero values, we can divide by x^4 (which will always be positive, so we don't have to worry about flipping the sign). Then the question becomes:
x^1 > 1? In other words, is x > 1 or is x = 0?

Often on inequalities DS questions, the easiest way to rephrase the question might be to ask: what kinds of numbers are larger when taken to the 5th than the 4th? What kinds of numbers aren't? We can use this number line to test cases:

Image

We can clearly see that the only time we get a "yes" answer is when x > 1, so our target question is: is x > 1 ?

(I couldn't initially tell if those were dashes or negative signs in your notation, so just be careful with that).

1) x^3 > -x
We can't divide here because these are odd exponents. Use the number line to test:

Image

As we can see, this will be true wherever x > 0, so the statement is really saying: x > 0. This doesn't help us to answer the question "is x > 1?" INSUFFICIENT

2) 1/x < x
Again, we can't divide. Let's test cases:

Image

This will be true when x is a negative fraction, or when x > 1. INSUFFICIENT

1 & 2) If we combine the statements, then x has to be positive, so x cannot be a negative fraction. It must be greater than 1. SUFFICIENT.
Ceilidh Erickson
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by ceilidh.erickson » Tue Aug 27, 2013 12:30 pm
For more on using number lines to test cases, see:
https://www.beatthegmat.com/x-x-x-which- ... tml#583371
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by Mike@Magoosh » Tue Aug 27, 2013 4:55 pm
faraz_jeddah wrote:Is x^5 > x^4?

1 - x^3 > -x
2 - 1/x < x

I want to use the algebraic approach here.
Don't use the algebraic approach. This question is about chunks of real estate on the real number line, and you are trying to figure out which chunks of the number line come into play.

See this blog:
https://magoosh.com/gmat/2012/exponent-p ... -the-gmat/

Mike :-)
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by faraz_jeddah » Tue Aug 27, 2013 11:01 pm
I thank the experts for their feedback.

However, I want to know what is wrong with the approach I have used, especially the parts I have highlighted in blue. Are those steps viable or not?

Similar questions I have based my approach on
Is x = 2?

1 - Some Quadratic equation that gives us x = 2 or -3 (not sufficient)
2 - Some Quadratic equation that gives us x = 3 or 2 (not sufficient)

Combining both x = 2. SUFFICIENT
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by faraz_jeddah » Wed Aug 28, 2013 8:39 am
ceilidh.erickson wrote: 1 & 2) If we combine the statements, then x has to be positive, so x cannot be a negative fraction. It must be greater than 1. SUFFICIENT.
I understood everything until the point above. How do you combine?

The overlaps of 2 "yes" ?
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by Mike@Magoosh » Wed Aug 28, 2013 11:00 am
faraz_jeddah wrote:Is x^5 > x^4?
1 - x^3 > -x
2 - 1/x < x
I want to use the algebraic approach here.

Target question can be rephrased as
Is x^4(x-1) > 0?

Which means either
x^4>0 and x>1 (can I summarize this as x>1?)
OR
x^4<0 and x<1 (Can I summarize this as x<0?)

So Now when I try to evaluate the statements, I should try to prove either x>1 or x<0
Yes, since (x^4) is always positive for all non-zero numbers, the real question comes down to "Is x > 1?" That's a crucially important simplification. Very good!
faraz_jeddah wrote: Statement 1
x^3 > - x
This can be true when x>1 and not true when x is a negative fraction.

Statement 2
1/x < x
This can be true when x>1 and not true when x is a positive fraction.

So when I combine the statements I find the 'common' condition is x>1 which is sufficient.

I want to know if this approach is correct or if I have overlooked anything.

Thanks.
Statement #1:
x^3 > - x
True for all positive numbers, both numbers larger than one and fractions between 0 and 1, since any positive is greater than any negative. Not true for anything negative, since the left side would be negative, the right side would be positive, and any negative is less than any positive.
This statement, by itself, is equivalent to the statement x > 0, which by itself, is not sufficient.

Statement #2:
1/x < x
This is true for all positive number greater than one.
This is false for positive fractions between 0 and 1
This is true for all negative fractions between 0 and -1.
This is false for all negative numbers less than -1 (i.e. those with absolute value greater than 1)
This statement is equivalent to: (x > 1) OR (0 > x > -1), which by itself is not sufficient.

Combined --- the overlap is the region (x > 1), which is sufficient.

You didn't specify everything that was possible or not possible in each individual statement, but perhaps you were focused purely on what would be relevant in the "combined" case. Perhaps you understand everything that was possible or not possible in each of the individual statements and just didn't choose to say it. If you have any questions on that, please let me know.

Mike :-)
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