Apr 20 07:00 AM PDT  09:00 AM PDT Christina scored 760 by having clear (ability) milestones and a trackable plan to achieve the same. Attend this webinar to learn how to build trackable milestones that leverage your strengths to help you get to your target GMAT score. Apr 20 10:00 PM PDT  11:00 PM PDT The Easter Bunny brings … the first day of school?? Yes! Now is the time to start studying for the GMAT if you’re planning to apply to Round 1 of fall MBA programs. Get a special discount with the Easter sale! Apr 21 07:00 AM PDT  09:00 AM PDT Get personalized insights on how to achieve your Target Quant Score. Apr 21 10:00 PM PDT  11:00 PM PDT $84 + an extra $10 off for the first month of EMPOWERgmat access. Train to be ready for Round 3 Deadlines with EMPOWERgmat's Score Booster. Ends April 21st Code: GCENHANCED Apr 22 08:00 AM PDT  09:00 AM PDT What people who reach the high 700's do differently? We're going to share insights, tips, and strategies from data we collected on over 50,000 students who used examPAL. Save your spot today! Apr 23 08:00 PM EDT  09:00 PM EDT Strategies and techniques for approaching featured GMAT topics. Tuesday, April 23rd at 8 pm ET Apr 24 08:00 PM EDT  09:00 PM EDT Maximize Your Potential: 5 Steps to Getting Your Dream MBA Part 3 of 5: Key TestTaking Strategies for GMAT. Wednesday, April 24th at 8 pm ET
Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 05 May 2011
Posts: 354
Location: United States (WI)
WE: Research (Other)

Does 4^a = 4^a + b? (1) 16^a = 1 + ((2^2a) / (b^1)) (2) a
[#permalink]
Show Tags
14 Sep 2011, 10:31
Question Stats:
56% (01:59) correct 44% (02:32) wrong based on 124 sessions
HideShow timer Statistics
Does 4^a = 4^a + b? (1) 16^a = 1 + ((2^2a) / (b^1)) (2) a = 2 Hello, I'm having trouble understanding why statement 1 is sufficient.
In the explanation, statement 1 is simplified to 4^2a = 1 + (4^a)b and then both sides are divided by 4^a.
Wouldn't 4^2a divided by 4^a be 4^2? Or what am I missing? How does it simplify in such a manner?
Thank you.
Official Answer and Stats are available only to registered users. Register/ Login.



GMAT Tutor
Joined: 24 Jun 2008
Posts: 1418

Re: MGMAT Advanced quant #416 (pg 118)
[#permalink]
Show Tags
14 Sep 2011, 10:59
rjdunn03 wrote: Wouldn't 4^2a divided by 4^a be 4^2? From the exponent rules, if you divide two numbers with exponents, you subtract the powers. So, for example, 2^9/2^4 = 2^(94) = 2^5. In the same way, 4^(2a)/4^a = 4^(2a  a) = 4^a; it is not equal to 4^2 (unless a is equal to 2). I'm just running out the door, so I don't even have time to actually look at the question, but I can later if you still aren't clear on the solution.
_________________
GMAT Tutor in Toronto
If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com



Senior Manager
Joined: 05 May 2011
Posts: 354
Location: United States (WI)
WE: Research (Other)

Re: MGMAT Advanced quant #416 (pg 118)
[#permalink]
Show Tags
14 Sep 2011, 11:19
IanStewart wrote: rjdunn03 wrote: Wouldn't 4^2a divided by 4^a be 4^2? From the exponent rules, if you divide two numbers with exponents, you subtract the powers. So, for example, 2^9/2^4 = 2^(94) = 2^5. In the same way, 4^(2a)/4^a = 4^(2a  a) = 4^a; it is not equal to 4^2 (unless a is equal to 2). I'm just running out the door, so I don't even have time to actually look at the question, but I can later if you still aren't clear on the solution. Oh yes, thank you. I understand now, I wasn't thinking it through properly. I appreciate your help.



Director
Status: My Thread Master Bschool Threads>Krannert(Purdue),WP Carey(Arizona),Foster(Uwashngton)
Joined: 28 Jun 2011
Posts: 815

Re: MGMAT Advanced quant #416 (pg 118)
[#permalink]
Show Tags
14 Sep 2011, 12:59
Statement 1 is only sufficient ....We can solve for a and b values....
Statement 2 in suff..
IMO A



Senior Manager
Joined: 23 Oct 2010
Posts: 347
Location: Azerbaijan
Concentration: Finance

Re: MGMAT Advanced quant #416 (pg 118)
[#permalink]
Show Tags
15 Sep 2011, 00:27
rjdunn03 wrote: Does 4^a = 4^a + b? Wouldn't 4^2a divided by 4^a be 4^2? Or what am I missing? How does it simplify in such a manner? Thank you. no it wouldnt. 4^2a/4^a= 4^(2aa)=4^a rjdunn03 wrote: Does 4^a = 4^a + b?
(1) 16^a = 1 + ((2^2a) / (b^1)) (2) a = 2
Thank you. we have 4^a = 4^a + b or 4^2a = 1+4^a*b stmt 1 16^a = 1 + ((2^2a) / (b^1)) 4^2a=1+4^a*b suff stmt 2 is insuf, cuz b could be anything OA is A
_________________
Happy are those who dream dreams and are ready to pay the price to make them come true
I am still on all gmat forums. msg me if you want to ask me smth



Director
Joined: 01 Feb 2011
Posts: 646

Re: MGMAT Advanced quant #416 (pg 118)
[#permalink]
Show Tags
15 Sep 2011, 16:59
4^a = 4^(a) + b?
rephrasing we have 4^a = (1/(4^a)) + b?
=> (4^a)(4^a) = 1 + (4^a)*b?
=> 16^a = 1 + (4^a)*b?
1. Sufficient
16^a = 1+ (4^a)/(b^1)
= 1+(4^a)*b
2. Not sufficient.
we dont know anything about b.
Answer is A.



Manager
Joined: 27 Aug 2014
Posts: 138
Concentration: Finance, Strategy
GPA: 3.9
WE: Analyst (Energy and Utilities)

Re: Does 4^a = 4^a + b? (1) 16^a = 1 + ((2^2a) / (b^1)) (2) a
[#permalink]
Show Tags
26 Nov 2014, 15:00
rjkaufman21 wrote: Does 4^a = 4^a + b? (1) 16^a = 1 + ((2^2a) / (b^1)) (2) a = 2 Hello, I'm having trouble understanding why statement 1 is sufficient.
In the explanation, statement 1 is simplified to 4^2a = 1 + (4^a)b and then both sides are divided by 4^a.
Wouldn't 4^2a divided by 4^a be 4^2? Or what am I missing? How does it simplify in such a manner?
Thank you. I did it in the following way: let \(4^a = t\), then the question becomes, is \(t^2tb1 = 0\)? from statement 1:\(t^2 = 1 + tb\), or\(t^2tb1 = 0\), the required info from statement 2: we cannot tell if \(t^2tb1 = 0\), so NSF Answer [A]



Manager
Joined: 07 Apr 2015
Posts: 162

Re: Does 4^a = 4^a + b? (1) 16^a = 1 + ((2^2a) / (b^1)) (2) a
[#permalink]
Show Tags
12 May 2015, 03:42
Is it correct to simplify (4^a)*(4^a) as (4^a)^2 and then as 4^2a which equals 16^a?



Math Expert
Joined: 02 Sep 2009
Posts: 54369

Re: Does 4^a = 4^a + b? (1) 16^a = 1 + ((2^2a) / (b^1)) (2) a
[#permalink]
Show Tags
12 May 2015, 03:52
noTh1ng wrote: Is it correct to simplify (4^a)*(4^a) as (4^a)^2 and then as 4^2a which equals 16^a? Yes. You could also do (4^a)*(4^a) = 4^(a+a) = 4^(2a) = 16^a.
_________________



Board of Directors
Joined: 17 Jul 2014
Posts: 2556
Location: United States (IL)
Concentration: Finance, Economics
GPA: 3.92
WE: General Management (Transportation)

Re: Does 4^a = 4^a + b? (1) 16^a = 1 + ((2^2a) / (b^1)) (2) a
[#permalink]
Show Tags
22 Nov 2016, 09:02
rjkaufman21 wrote: Does 4^a = 4^a + b? (1) 16^a = 1 + ((2^2a) / (b^1)) (2) a = 2 Hello, I'm having trouble understanding why statement 1 is sufficient.
In the explanation, statement 1 is simplified to 4^2a = 1 + (4^a)b and then both sides are divided by 4^a.
Wouldn't 4^2a divided by 4^a be 4^2? Or what am I missing? How does it simplify in such a manner?
Thank you. B is clearly insufficient. we can rewrite 4^a = 4^a + b as: 2^2a = 2^2a +b or (2^4a  1)/2^2a = b 1. says exactly this. (2^2a) / (b^1) = 2^2a * b 16^a = 2^4a 2^4a  1 = 2^2a*b b= (2^4a  1)/2^2a sufficient.



NonHuman User
Joined: 09 Sep 2013
Posts: 10543

Re: Does 4^a = 4^a + b? (1) 16^a = 1 + ((2^2a) / (b^1)) (2) a
[#permalink]
Show Tags
19 Mar 2019, 00:26
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________




Re: Does 4^a = 4^a + b? (1) 16^a = 1 + ((2^2a) / (b^1)) (2) a
[#permalink]
19 Mar 2019, 00:26






