GreenlightTestPrep wrote:
x, y and p are integers, and xyp ≠ 0. If \(p^x < p^y\), which of the following MUST be true?
i) \(x - y < 0\)
ii) \(x < 2y\)
iii) \(x^p < y^p\)
A) i only
B) ii only
C) iii only
D) i and ii only
E) none of the above
Two important rules:
ODD exponents preserve the sign of the base. So, (
NEGATIVE)^(
ODD integer) =
NEGATIVEand (
POSITIVE)^(
ODD integer) =
POSITIVEAn EVEN exponent always yields a positive result (unless the base = 0)
So, (
NEGATIVE)^(
EVEN integer) =
POSITIVEand (
POSITIVE)^(
EVEN integer) =
POSITIVE------------------------------------
So, one solution to the inequality \(p^x < p^y\) is \(p = -1\), \(x = 7\) and \(y = 2\)
Plugging those values into the inequality, we get: \((-1)^7 < (-1)^2\)
Simplify to get: \(-1 < 1\), WORKS.
Now plug \(p = -1\), \(x = 7\) and \(y = 2\) into the three statements to get:
i) \(7 - 2 < 0\)
Simplify to get: \(5 < 0\)
NOT true.
So,
statement i need not be true.
ii) \(7 < 2(2)\)
Simplify to get: \(7 < 4\)
NOT true.
So,
statement ii need not be true.
iii) \(7^{-1} < 2^{-1}\)
Simplify to get: \(\frac{1}{7} < \frac{1}{2}\)
This is TRUE.
So, we can't (yet) conclude that
statement iii need not be true.
-------------------------------------
Let's see if any other values will show that statement iii need not be true.
Another solution to the inequality \(p^x < p^y\) is \(p = -1\), \(x = 1\) and \(y = 2\)
Plugging those values into the inequality, we get: \((-1)^1 < (-1)^2\)
Simplify to get: \(-1 < 1\), WORKS.
Now plug \(p = -1\), \(x = 1\) and \(y = 2\) into statement iii to get:
iii) \(1^{-1} < 2^{-1}\)
Simplify to get: \(\frac{1}{1} < \frac{1}{2}\)
NOT true.
So,
statement iii need not be true.
Answer: E
Cheers,
Brent
_________________
Brent Hanneson – Creator of greenlighttestprep.comSign up for GRE Question of the Day emails