GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
x^3>x^5
[#permalink]
22 Jan 2024, 00:17
22 Jan 2024, 00:17
Expert Reply
1
Bookmarks
00:00
Question Stats:
60% (01:09) correct
40% (00:31) wrong based on 5 sessions
Hide Show timer Statistics
\(x^3>x^5\)
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
x |
0 |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Part of the project: The Butler-GRE Daily New Quant and Verbal Questions to Practice (2023) - Gain 20 Kudos & Get FREE Access to GRE Prep Club TESTS
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Free Materials for the GRE General Exam - Where to get it!!
GRE Geometry Formulas
GRE - Math Book
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Free Materials for the GRE General Exam - Where to get it!!
GRE Geometry Formulas
GRE - Math Book
Manager
Joined: 11 Nov 2023
Posts: 203
Given Kudos: 76
WE:Business Development (Advertising and PR)
Re: x^3>x^5
[#permalink]
22 Jan 2024, 13:01
22 Jan 2024, 13:01
1
This is testing knowledge of negative numbers and fractions as bases of exponents.
Let x = -2 < 0.
\(x^5 = (-2)^5 = -32\)
\(x^3 = (-2)^3 = -8\)
\(-8 > -32\), which satisfies \(x^3>x^5\).
Let x = 1/2 > 0.
\((\frac{1}{2})^5 = \frac{1}{32}\)
\((\frac{1}{2})^3 = \frac{1}{8}\)
\(\frac{1}{8} > \frac{1}{32}\), which also satisfies x^3>x^5.
Rule of thumb with tackling comparisons of exponents is to consider negatives AND fractions, on top of positives integers:
- If the base is a fraction: the greater the power, the closer to 0 it will get
- If the base is negative: the greater the odd power, the further from 0 it will get
Let x = -2 < 0.
\(x^5 = (-2)^5 = -32\)
\(x^3 = (-2)^3 = -8\)
\(-8 > -32\), which satisfies \(x^3>x^5\).
Let x = 1/2 > 0.
\((\frac{1}{2})^5 = \frac{1}{32}\)
\((\frac{1}{2})^3 = \frac{1}{8}\)
\(\frac{1}{8} > \frac{1}{32}\), which also satisfies x^3>x^5.
Rule of thumb with tackling comparisons of exponents is to consider negatives AND fractions, on top of positives integers:
- If the base is a fraction: the greater the power, the closer to 0 it will get
- If the base is negative: the greater the odd power, the further from 0 it will get
gmatclubot
Re: x^3>x^5 [#permalink]
22 Jan 2024, 13:01
Moderators:
