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

If x and y are nonzero integers, is \(x^y < y^x\)?

(1) \(x = y^2\) --> if \(x=y=1\), then \(x^y=1=y^x\), so the answer would be NO BUT if \(y=3\) and \(x=9\), then \(x^y=9^3<y^x=3^9\), so the answer would be YES. Not sufficient.

(2) \(y>2\). No info about \(x\), not sufficient.

(1)+(2) From (1) \(x = y^2\), thus the question becomes: is \((y^2)^y<y^{(y^2)}\)? --> is \(y^{2y}<y^{(y^2)}\)? Now, since from (2) \(y=integer>2\), then \(2y\) will always be less than \(y^2\), therefore \(y^{2y}\) will be less than \(y^{(y^2)}\). Sufficient.

Answer: C.




If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus:
\(a^m^n=a^{(m^n)}\) and not \((a^m)^n\), which on the other hand equals to \(a^{mn}\).

So:
\((a^m)^n=a^{mn}\);

\(a^m^n=a^{(m^n)}\) and not \((a^m)^n\).

Hope it's clear.
_________________

Target question: Is x^y < y^x ?

Given: x and y are nonzero integers

Statement 1: x = y²
Take the original target question, and replace x with y² to get: Is (y²)^y < y^(y²)?
Simplify the left side power to get: Is y^(2y) < y^(y²)?

Let's TEST some values of y.
Case a: If y = 1, then y^(2y) = 1 and y^(y²) = 1. In this case, the answer to the target question is NO, it is NOT the case that y^(2y) < y^(y²)
Case b: If y = 3, then y^(2y) = 3^6 and y^(y²) = 3^9. In this case, the answer to the target question is YES, it IS the case that y^(2y) < y^(y²)
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: y > 2
Since we have no information about x, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
From statement 1, we changed the target question to read Is y^(2y) < y^(y²)?
Now let's divide both sides of the inequality by y^(2y) to get: Is 1 < [y^(y²)/y^(2y)]?
Apply the Quotient Law to get: Is 1 < y^(y² - 2y)?
Notice that, if y is an INTEGER greater than 2 (i.e., statement 2), then y² - 2y is greater than 1
So, y^(y² - 2y) MUST be greater than 1
So, the answer to the target question is YES it IS the case that 1 < y^(y² - 2y)
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer: C

Cheers,
Brent

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.

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?

Please enlighten. Thanks.

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.
_________________

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.

Therefore, C is the answer.

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.

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

Hence answer is Option C

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.
_________________

Thanks. That makes complete sense.