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# If x and y are nonzero integers, is x^y < y^x ?

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Re: If x and y are nonzero integers, is x^y < y^x ? (1) x = y^2 (2) y > 2 [#permalink]
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This is a tricky question. I think it relies on you misapplying the rule: $$(x^a)^b = x^{ab}$$. Is this only valid if a and b are constants?

Example:
(1) $$x = y^2$$;

$$x^y < y^x$$ therefore, $$(y^2)^y < y^{y^2}$$. How do you simplify this? The guide shows to $$y^{2y} < y^{y^2}$$. The left hand side makes sense to me.

Why would $$y^{y^2}$$ not simplify to $$y^{2y}$$ also? Plugging in numbers, it makes sense. I just want to understand the concept.
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Re: If x and y are nonzero integers, is x^y < y^x ? (1) x = y^2 (2) y > 2 [#permalink]
Posting this msg here even though i sent a private msg to you-for the benefit of others here.

Hi Bunuel, apprecite ur wonderful explanation. I am having trouble in DS question where x & y are termed as non-zero integers.

What is the best way to analyze instances where x & y are are NEGATIVE integers. I see that u have not analyzed this possibility. is there a trick to be sure that this is not needed as u have solved in this case?

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Re: If x and y are nonzero integers, is x^y < y^x ? (1) x = y^2 (2) y > 2 [#permalink]
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ramanankris
Posting this msg here even though i sent a private msg to you-for the benefit of others here.

Hi Bunuel, apprecite ur wonderful explanation. I am having trouble in DS question where x & y are termed as non-zero integers.

What is the best way to analyze instances where x & y are are NEGATIVE integers. I see that u have not analyzed this possibility. is there a trick to be sure that this is not needed as u have solved in this case?

On DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another.

For statement (1) I got YES answer and then NO answer with positive numbers, so my goal to prove that this statement was not sufficient was reached, hence there was no need to try negative numbers.

Hope it's clear.
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Re: If x and y are nonzero integers, is x^y < y^x ? [#permalink]
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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 2 variables ($$x$$ and $$y$$) and 0 equations, C is most likely to be the answer. So, we should consider 1) & 2) first.

Conditions 1) & 2):
$$x^y = (y^2)^y = y^{2y}$$
$$y^x = y^{y^2}$$
Since $$y > 2$$, we have $$y^2 > 2y$$ and $$y^{2y} < y^{y^2}$$.
Thus both conditions together are sufficient.

Since this is an integer question (one of the key question areas), we should also consider choices A and B by CMT(Common Mistake Type) 4(A).

Condition 1)
$$x = 9, y = 3 ⇒ 9^3 = 3^6 < 3^9$$ : Yes
$$x = 1, y = 1 ⇒ 1^1 = 1^1$$ : No
The condition 1) only is not sufficient.

Condition 2)
$$x = 1, y = 4 ⇒ 1^2 < 4^1$$ : Yes
$$x = -1, y = 4 => (-1)^4 = 1 > 4^{-1} = \frac{1}{4}$$ : No
The condition 2) only is not sufficient.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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Re: If x and y are nonzero integers, is x^y < y^x ? [#permalink]
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From Statement 1 x=y^2 so x and y both can be equal to 1 , then x^y cannot be less than y^x as both are equal
When y=3 then x=9 then x^y will be less than y^x. So no unique answer. Hence statement 1 alone is not suffice

Statement 2, y>2 does not provides us with any data about x hence not suffice as we cannot compare the expressions

Both statements together are required. When we combine both the statements, only value possible is y=3 and x=9
Hence Option C

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If x and y are nonzero integers, is x^y < y^x ? [#permalink]
Bunuel

Probably I am quite sleepy, but I am struggling to prove "NO" with statement 2.
If y>2, whatever value of x I take I get a yes.
Can you share a set of numbers where we can get a no for this?

Pls Ignore: I finally figured. Just posting in case someone else may find it useful.
I was just testing Y = 3, and tried all sorts of values for X. Y = 4, and X = -2 did the trick.
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If x and y are nonzero integers, is x^y < y^x ? [#permalink]
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If x and y are nonzero integers, is x^y < y^x ?

(1) x = y^2
(2) y > 2

Bunuel

Probably I am quite sleepy, but I am struggling to prove "NO" with statement 2.
If y>2, whatever value of x I take I get a yes.
Can you share a set of numbers where we can get a no for this?

Pls Ignore: I finally figured. Just posting in case someone else may find it useful.
I was just testing Y = 3, and tried all sorts of values for X. Y = 4, and X = -2 did the trick.

To get $$x^y < y^x$$, probably easiest would be to try negative x, for example: x = - 1 and y = ANY even number greater than 2.

However, there are also examples where x is positive:

x = 2 and y = 5, 6, 7, 8, 9, or 10;

x = 3 and y = 4, 5, 6, 7, 8, 9, or 10;

x = 4 and y = 5, 6, 7, 8, 9, or 10;

x = 5 and y = 6, 7, 8, 9, or 10;

x = 6 and y = 7, 8, 9, or 10;

x = 7 and y = 8, 9, or 10;

x = 8 and y = 9.

Hope it helps.
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Re: If x and y are nonzero integers, is x^y < y^x ? [#permalink]
Thanks. That makes complete sense.
Re: If x and y are nonzero integers, is x^y < y^x ? [#permalink]
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