Find all School-related info fast with the new School-Specific MBA Forum

It is currently 01 Oct 2014, 12:35

Close

GMAT 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

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Tricky Exponents Question

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Current Student
User avatar
Status: Current MBA Student
Joined: 19 Nov 2009
Posts: 129
Concentration: Finance, General Management
GMAT 1: 720 Q49 V40
Followers: 8

Kudos [?]: 67 [0], given: 210

Tricky Exponents Question [#permalink] New post 13 Apr 2011, 12:23
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

33% (00:00) correct 67% (00:00) wrong based on 4 sessions
Can someone please walk me through the simplification steps on this problem. I got lost along the way. Thanks!

If 2^{x+2} - 3(2^{x-1})=5/8(2^{16-x), what is the value of x?

a. -7
b. 6
c. 7
d. 13
e. 16
2 KUDOS received
GMAT Instructor
avatar
Joined: 24 Jun 2008
Posts: 977
Location: Toronto
Followers: 259

Kudos [?]: 694 [2] , given: 3

Re: Tricky Exponents Question [#permalink] New post 13 Apr 2011, 14:18
2
This post received
KUDOS
tonebeeze wrote:
Can someone please walk me through the simplification steps on this problem. I got lost along the way. Thanks!

If 2^{x+2} - 3(2^{x-1})=5/8(2^{16-x), what is the value of x?

a. -7
b. 6
c. 7
d. 13
e. 16


There are two general principles that you can apply to any similar problem (a problem involving numbers and powers on either side of an equation):

1) get prime bases everywhere. Your goal is to get the same base on each side of the equation, in which case the exponents must be equal
2) if you are adding or subtracting two (or more) terms with the same base, factor out the *smaller* power


We can apply principle 2) on the left side of the equation in the question above, though it isn't as straightforward as in many other questions. If you were to see, for example:

7^7 + 7^5

you'd normally factor out the smaller power, which is the 7^5 term:

7^7 + 7^5 = 7^5 (7^2 + 1) = 7^5(50) = (2)(5^2)(7^5)

Notice that to get the two terms in the brackets (the 7^2 + 1), we just reduce the exponents by 5, because we factored out 7^5.

In the question above, we have an unknown in our power, but the same principle applies. If you see something like:

2^{x+2} - 2^{x-1}

we can again factor out the smaller power, which is 2^(x-1), to get

2^{x-1} (2^3 - 1)

Notice we are just subtracting x-1 from each power to work out the exponents on the two terms in the brackets.

Now, I think the way you've written the question might be confusing to some, because it appears that the 2^{16 - x} on the right side of the equation is in the denominator of a fraction. That can't be the case; then the equation cannot possibly be true. So I'm guessing the question ought to look like this:

\begin{align}
2^{x+2} - 3(2^{x-1}) &= \left(\frac{5}{8} \right) (2^{16-x}) \\
2^{x-1} (2^3 - 3) &= \left( \frac{5}{2^3} \right) (2^{16 - x}) \\
(2^{x-1})(5) &= (5)(2^{13 - x}) \\
2^{x-1} &= 2^{13 - x} \\
x - 1 &= 13 - x \\
x &= 7
\end{align}

_________________

Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

Private GMAT Tutor based in Toronto

Director
Director
avatar
Joined: 01 Feb 2011
Posts: 770
Followers: 14

Kudos [?]: 82 [0], given: 42

Re: Tricky Exponents Question [#permalink] New post 13 Apr 2011, 18:27
Going forward please format the question properly.

Right hand side expression appears as if its 5/(8* 2^(16-x)) where as its (5/8)(2^(16-x))

2^(x+2) - 3*2^(x-1) = (5/8) (2^(16-x))

=> 2^x *(5/2) = 5* 2^(13-x)

equating two powers on both sides we have

x-1 = 13-x

=> x =7

Answer is C.
SVP
SVP
avatar
Joined: 16 Nov 2010
Posts: 1691
Location: United States (IN)
Concentration: Strategy, Technology
Followers: 30

Kudos [?]: 291 [0], given: 36

Premium Member Reviews Badge
Re: Tricky Exponents Question [#permalink] New post 13 Apr 2011, 18:34
2^x(2^2 - 3*2^x/2) = 5/8*(2^(16-x))

2^x(4 * 2 - 3)/2 = 5/8 * 2^16/2^x

=> 2^2x (5/2) = 5/2^3 * 2^16

=> 2^2x = (2)^16 + 1 - 3

=> 2^2x = 2^14

=> 2x = 14

=> x = 14/2 = 7

Answer - C
_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Manager
Manager
avatar
Joined: 17 Jan 2011
Posts: 243
Followers: 1

Kudos [?]: 22 [0], given: 4

Re: Tricky Exponents Question [#permalink] New post 15 Apr 2011, 19:56
Select a random value from choices, lets say x=6
=> 2^8 - 3.2^5 = (5/8).2^10
=> 2^5(2^3 - 3) = (5).2^7
=> 2^5(5) = (5).2^7
the LHS should give a higher value and RHS lower to balance it. So we should try the next one with a slightly higher value
Answer - C
_________________

Good Luck!!!

***Help and be helped!!!****

Manager
Manager
avatar
Joined: 05 Jul 2010
Posts: 205
Followers: 1

Kudos [?]: 10 [0], given: 18

Re: Tricky Exponents Question [#permalink] New post 16 Apr 2011, 14:32
good explanation! IanStewart. +1
Intern
Intern
avatar
Joined: 06 Apr 2011
Posts: 46
Followers: 0

Kudos [?]: 5 [0], given: 4

GMAT ToolKit User
Re: Tricky Exponents Question [#permalink] New post 16 Apr 2011, 20:15
Upon simplification:
2^{x+2} * (5/8) = 2^{16-x} * (5/8)

which implies x+2 = 16-x,
x = 7.

Answer C
Re: Tricky Exponents Question   [#permalink] 16 Apr 2011, 20:15
    Similar topics Author Replies Last post
Similar
Topics:
118 Experts publish their posts in the topic NEW!!! Tough and tricky exponents and roots questions Bunuel 70 12 Jan 2012, 02:50
211 Experts publish their posts in the topic NEW!!! Tough and tricky exponents and roots questions Bunuel 151 12 Jan 2012, 02:03
Tricky question vix 2 23 Nov 2011, 05:28
1 A tricky question about Exponents Lstadt 4 26 Jul 2011, 11:31
Tricky Exponents Question lfox2 3 27 Jun 2006, 02:21
Display posts from previous: Sort by

Tricky Exponents Question

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.