Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
26 Jun 2010, 19:14
Kudos
Bookmarks
N
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
(N/A)
Question Stats:
67% (00:37) correct
33%
(00:58)
wrong
based on 4
sessions
History
Date
Time
Result
Not Attempted Yet
Solve for x:
2^x - 2^(x-2) = 3(2^13)
OA:
how do you solve this?
2^x - 2^(x-2) = 3(2^13)
OA:
x = 15
how do you solve this?
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
26 Jun 2010, 20:19
Kudos
Bookmarks
Hi,
had you included the answer choices, an approach quicker than algebra may have presented itself....
2^x - 2^(x-2) = 3(2^13)
On the left we see the second term has the "-" sign in the exponent. So, we should recall that we subtract exponents when dividing:
2^x - 2^x/2^2 = 3(2^13)
Multiply each term by 2^2 to get rid of the fraction
(2^x)(2^2) - 2^x = 3(2^13)(2^2)
Factor out 2^x on the left and add "13" and "2" on the right
2^x(2^2 - 1) = 3(2^15)
2^x(3) = 3(2^15)
3(2^x) = 3(2^15)
Thus, x = 15
had you included the answer choices, an approach quicker than algebra may have presented itself....
2^x - 2^(x-2) = 3(2^13)
On the left we see the second term has the "-" sign in the exponent. So, we should recall that we subtract exponents when dividing:
2^x - 2^x/2^2 = 3(2^13)
Multiply each term by 2^2 to get rid of the fraction
(2^x)(2^2) - 2^x = 3(2^13)(2^2)
Factor out 2^x on the left and add "13" and "2" on the right
2^x(2^2 - 1) = 3(2^15)
2^x(3) = 3(2^15)
3(2^x) = 3(2^15)
Thus, x = 15
whiplash2411
Joined: 09 Jun 2010
Last visit: 02 Mar 2015
Posts: 1,761
Own Kudos:
Given Kudos: 210
Status:Three Down.
Concentration: General Management, Nonprofit
Schools: HBS '17 (M) Stanford '17 (A)
26 Jun 2010, 21:34
Kudos
Bookmarks
\(2^x - 2^(x-2) = 3(2^13)\)
\(2^x - \frac{2^x}{4} = 3*(2^13)\)
\(2^x (1 - 1/4) = 3*(2^13)\)
\((2^x) ( 3/4)= 3*(2^13)\)
\(2^(x-2) * 3 = 3* (2^13)\)
\(x-2 = 13\)
\(x=15\)
Hope this helps!
Note: I don't know why it formats the powers this way, but the powers are generally 13 even though it doesn't look like that. And in the first line, it's (x-2)
\(2^x - \frac{2^x}{4} = 3*(2^13)\)
\(2^x (1 - 1/4) = 3*(2^13)\)
\((2^x) ( 3/4)= 3*(2^13)\)
\(2^(x-2) * 3 = 3* (2^13)\)
\(x-2 = 13\)
\(x=15\)
Hope this helps!
Note: I don't know why it formats the powers this way, but the powers are generally 13 even though it doesn't look like that. And in the first line, it's (x-2)
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
