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 > 3^20
[#permalink]
22 Nov 2023, 03:05
22 Nov 2023, 03:05
Expert Reply
1
Bookmarks
00:00
Question Stats:
60% (01:50) correct
40% (01:24) wrong based on 10 sessions
Hide Show timer Statistics
\(x > 3^{20}\)
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 |
\(x\) |
\(6^{12}\) |
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: 207
Given Kudos: 76
WE:Business Development (Advertising and PR)
x > 3^20
[#permalink]
01 Dec 2023, 20:03
01 Dec 2023, 20:03
1
Not 100% confident that this is the best approach, so correct me if there's a better way but:
I'd compare \(3^{20}\) to \(6^{12}\).
\(6^{12} = (3*2)^{12} = 3^{12}*2^{12}\)
Simplify to then compare \(3^8\) to \(2^{12}\).
This is where I'd brute-force calculate to get \(3^8=6561\) and \(2^{12}=4096\).
Now we know that \(3^{20}>6^{12}\) and \(x > 3^{20}\), x must be greater than \(6^{12}\).
I'd compare \(3^{20}\) to \(6^{12}\).
\(6^{12} = (3*2)^{12} = 3^{12}*2^{12}\)
Simplify to then compare \(3^8\) to \(2^{12}\).
This is where I'd brute-force calculate to get \(3^8=6561\) and \(2^{12}=4096\).
Now we know that \(3^{20}>6^{12}\) and \(x > 3^{20}\), x must be greater than \(6^{12}\).
Re: x > 3^20
[#permalink]
08 Jan 2024, 02:23
08 Jan 2024, 02:23
Carcass can you guide me? not able to figure out less time consuming way to solve this question
