Last visit was: 19 Jul 2025, 20:41 It is currently 19 Jul 2025, 20:41
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.
Close
Request Expert Reply
Confirm Cancel
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Jul 2025
Posts: 102,627
Own Kudos:
Given Kudos: 98,235
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,627
Kudos: 742,808
 [72]
1
Kudos
Add Kudos
71
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Jul 2025
Posts: 102,627
Own Kudos:
742,808
 [6]
Given Kudos: 98,235
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,627
Kudos: 742,808
 [6]
2
Kudos
Add Kudos
4
Bookmarks
Bookmark this Post
General Discussion
User avatar
TestPrepUnlimited
Joined: 17 Sep 2014
Last visit: 30 Jun 2022
Posts: 1,226
Own Kudos:
1,067
 [16]
Given Kudos: 6
Location: United States
GMAT 1: 780 Q51 V45
GRE 1: Q170 V167
Expert
Expert reply
GMAT 1: 780 Q51 V45
GRE 1: Q170 V167
Posts: 1,226
Kudos: 1,067
 [16]
4
Kudos
Add Kudos
12
Bookmarks
Bookmark this Post
User avatar
yashikaaggarwal
User avatar
Senior Moderator - Masters Forum
Joined: 19 Jan 2020
Last visit: 17 Jul 2025
Posts: 3,095
Own Kudos:
Given Kudos: 1,510
Location: India
GPA: 4
WE:Analyst (Internet and New Media)
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Answer is supposedly B.
A. 2^4/3= 2.51 < x < 2^3/2= 2.82 (less than 2^x=3)

B. 2^3/2 =2.82 < x < 2^5/3= 3.17 (3 lies in between)

C. 2^5/3= 3.17 < x < 2^7/4= 3.36 (More than 2^x=3)

D. 2^7/4= 3.36 < x < 2^11/6= 3.56 (More than 2^x=3)

E. 2^11/6= 3.56 < x < 2^2= 4 (More than 2^x=3)

B is the answer

Posted from my mobile device
User avatar
nick1816
User avatar
Retired Moderator
Joined: 19 Oct 2018
Last visit: 28 Jun 2025
Posts: 1,853
Own Kudos:
7,843
 [1]
Given Kudos: 707
Location: India
Posts: 1,853
Kudos: 7,843
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
\(x = log_2 3\)

\(x= \frac{log_{10}3}{log_{10}2}\)

\(x= \frac{0.47}{0.3} \)

\(1.6>\frac{0.47}{0.3}>1.5\)

OR

\(2^{1.5} = 2^{(1+0.5)} = 2*2^{0.5} = 2*1.414 = 2.82\)

\(2^{1+\frac{2}{3}} = 2*2^{\frac{2}{3}} = 2*4^{\frac{1}{3}} = 2*[\frac{500}{125}]^{\frac{1}{3}} \) ~ \(2*[\frac{512}{125}]^{\frac{1}{3}}\)= \(2*(\frac{8}{5}) = 3.2\)



Or

\((2^3)*(\frac{9}{8}) = 3^2\)

\(2^3*1.125 = 3^2\)

We know that 2^0= 1 and 2^1 = 2

1.125 must be in between \(2^{0.1}\) and \(2^{0.2}\)

\(2^3*2^{0.1} < 2^{2x} < 2^3*2^{0.2}\)

\(2^{3.1} < 2^{2x} < 2^{3.2}\)

\(2^{1.55}< 2^x< 2^{1.6}\)

B






Bunuel
If \(2^x = 3\), then which of the following must be true?


A. \(1 \frac{1}{3} < x < 1 \frac{1}{2}\)

B. \(1 \frac{1}{2} < x < 1 \frac{2}{3}\)

C. \(1 \frac{2}{3} < x < 1 \frac{3}{4}\)

D. \(1 \frac{3}{4} < x < 1 \frac{5}{6}\)

E. \(1 \frac{5}{6} < x < 2\)

Are You Up For the Challenge: 700 Level Questions
User avatar
AnirudhaS
User avatar
LBS Moderator
Joined: 30 Oct 2019
Last visit: 25 Jun 2024
Posts: 812
Own Kudos:
Given Kudos: 1,575
Posts: 812
Kudos: 850
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I solved this by "feeling". Let me explain. But I need experts to critic me on this method. nick1816 let me know your thoughts please?

So we know
2^1.1
2^1.2
2^1.3
.....
2^1.9
2^2

is going to be an exponential curve (NOT linear).

And if it was linear then 2^1.5 would be exactly 3, but its not.
So 2^1.5 < 3
But the rate of growth of this exponential curve is only slight (not as much as say 2^2, 2^3, 2^4...)
So 2^1.5 should be quite close to 3

I mean then I chose the option B which is closest. However, I do understand that this is not full proof (as we do not know how close 2^1.5 is to 3 unless we work it out)
User avatar
nick1816
User avatar
Retired Moderator
Joined: 19 Oct 2018
Last visit: 28 Jun 2025
Posts: 1,853
Own Kudos:
7,843
 [1]
Given Kudos: 707
Location: India
Posts: 1,853
Kudos: 7,843
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
You are on point bhai. Since the curve is exponential, 2^1.5 < 3.

Now you could find the approximate value of \(2^{(1\frac{2}{3})}\) just to make sure that you're marking the correct option. I've already mentioned one way to approximate \(2^{(1\frac{2}{3})}\)in my solution.

Another way is-

\(3375 < 4000 < 4096\)

\(15^3<4000<16^3\)

\(1.5^3 < 4 < 1.6^3\)

\(1.5 < 4^{(\frac{1}{3})} < 1.6\)

\(1.5 < 2^{(\frac{2}{3})} < 1.6\)

\(1.5*2 < 2*2^{(\frac{2}{3})} < 1.6*2\)

\(3 < 2^1*2^{(\frac{2}{3})} < 3.2\)

\(3 < 2^{(1+\frac{2}{3})} < 3.2\)



AnirudhaS
I solved this by "feeling". Let me explain. But I need experts to critic me on this method. nick1816 let me know your thoughts please?

So we know
2^1.1
2^1.2
2^1.3
.....
2^1.9
2^2

is going to be an exponential curve (NOT linear).

And if it was linear then 2^1.5 would be exactly 3, but its not.
So 2^1.5 < 3
But the rate of growth of this exponential curve is only slight (not as much as say 2^2, 2^3, 2^4...)
So 2^1.5 should be quite close to 3

I mean then I chose the option B which is closest. However, I do understand that this is not full proof (as we do not know how close 2^1.5 is to 3 unless we work it out)
User avatar
satya2029
Joined: 10 Dec 2017
Last visit: 19 Jul 2025
Posts: 231
Own Kudos:
Given Kudos: 137
Location: India
Posts: 231
Kudos: 242
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
If \(2^x = 3\), then which of the following must be true?


A. \(1 \frac{1}{3} < x < 1 \frac{1}{2}\)

B. \(1 \frac{1}{2} < x < 1 \frac{2}{3}\)

C. \(1 \frac{2}{3} < x < 1 \frac{3}{4}\)

D. \(1 \frac{3}{4} < x < 1 \frac{5}{6}\)

E. \(1 \frac{5}{6} < x < 2\)

Are You Up For the Challenge: 700 Level Questions

2^x=3
2^x-2=1
2^(x-1)=3/2=1.5
\(\sqrt{2}=1.414\)
0.5<x<0.6
B:)
User avatar
Bambi2021
Joined: 13 Mar 2021
Last visit: 23 Dec 2021
Posts: 321
Own Kudos:
Given Kudos: 226
Posts: 321
Kudos: 131
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Must admit I failed this question. But I found some valuable approximations of x. The fractions represent the closest distances between a power of 2 and a power of 3.

https://oeis.org/A254351/internal

"log(3)/log(2) = 1.5849625... is an irrational number. The fractions (2/1, 3/2, 5/3, 8/5, 11/7, 19/12, 46/29, 65/41, 84/53, 317/200, 401/253, 485/306, 569/359, 1054/665, ...) are a sequence of approximations to log(3)/log(2), where each is an improvement on its predecessors."

This means that 2^(19/12) is closer to 3 than 2^(11/7).

"Seven/eleven" seems easy to remember as a decent approximation. =)

Posted from my mobile device
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 37,453
Own Kudos:
Posts: 37,453
Kudos: 1,013
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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.
Moderators:
Math Expert
102627 posts
PS Forum Moderator
698 posts