BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

sum of unknown power

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 103
Joined: Wed May 30, 2007 6:17 amFollowed by: 1 member
by tanyajoseph » Fri Aug 24, 2007 12:10 am
If s, u, and v are positive integers and 2^s = 2^u + 2^v, which of the following must be true?

I. s= u
II. u not equal to v
III. s>v

A) None
B) I only
C) II only
D) III only
E) II and III
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Senior | Next Rank: 100 Posts
Posts: 65
Joined: Tue Jun 12, 2007 12:57 am

by erdnah » Fri Aug 24, 2007 1:17 am
2^s = 2^u + 2^v
s = u + v

I. no
II. no
III. yes, s = v - u, and u is a positive integer and so cannot be 0!

D
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 101
Joined: Tue Aug 07, 2007 1:45 amFollowed by: 1 member

by agps » Fri Aug 24, 2007 2:07 am
I. if s=u then 2^v = 0, that is never possible, No
II. u not equal to v, we don't know s, so u can be equal to v for what we know, No
III. s>v s=u+v u and v (and s) are positive integers, so u+v is always > v, Yes

answer is D
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 233
Joined: Mon Jun 09, 2008 1:30 am
Thanked: 5 times

by blaster » Mon Jul 06, 2009 3:22 am
it's not correct s=u+v . you can't just sum all these figures

can someone give an explanation to the question? i can't solve that one. for me it must be A
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 15
Joined: Tue Jul 22, 2008 6:52 am
Thanked: 2 times

by Ad_Astra_Per_Aspera » Mon Jul 06, 2009 7:24 am
you could just plug numbers and check:

put s = 3 then we have 2^s = 8 , so in order for 2^u + 2^v be equal to 8 u and v can take the value of 2 each since 2^2 is 4 and so sum will be 8.

From the above it follows that when s = 3 , u =2 and v=2

now look at the statements ... only III holds true.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 527
Joined: Thu May 01, 2008 12:06 am
Thanked: 7 times

Re: sum of unknown power

by real2008 » Mon Jul 06, 2009 11:42 pm
tanyajoseph wrote:If s, u, and v are positive integers and 2^s = 2^u + 2^v, which of the following must be true?

I. s= u
II. u not equal to v
III. s>v

A) None
B) I only
C) II only
D) III only
E) II and III
if the answer is D, then read further....

2^s=2^u+2^v

or, (2^s)/(2^v)=(2^u)/(2^v)+1

or, 2^(s-v)= 2^(u-v)+1

or 2^(s-v)>2^(u-v)

taking log to both side,

(s-v)* log2> (u-v)* log2

or, s-v> u-v

or s>u

case I not true

case II

take s=2, u=1, v=1

CaseIII

s>v can be proven similarly as shown above.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1