Last visit was: 21 Apr 2026, 06:54 It is currently 21 Apr 2026, 06:54
Home
Messages
Tests
Schools
Events
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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
655-705 (Hard)|   Algebra|   Exponents|      
User avatar
niteshwaghray
User avatar
Joined: 14 Sep 2015
Last visit: 11 May 2018
Posts: 59
Own Kudos:
469
 [21]
Given Kudos: 19
Location: India
Schools: Jones '20 (WL) WP Carey" 20 (D) Boston U '20 (WL) Olin '20 (S) ISB '19 (D) Simon '20 (WL)
GMAT 1: 700 Q45 V40
GPA: 3.41
Schools: Jones '20 (WL) WP Carey" 20 (D) Boston U '20 (WL) Olin '20 (S) ISB '19 (D) Simon '20 (WL)
GMAT 1: 700 Q45 V40
Posts: 59
Kudos: 469
 [21]
If x and y are distinct positive integers, such that (2^x)(16^(x-1)) = Updated on: 25 May 2017, 14:08
Originally posted by niteshwaghray on 25 May 2017, 14:00.
Last edited by Bunuel on 25 May 2017, 14:08, edited 1 time in total.
Renamed the topic and edited the question.
4
Kudos
Add Kudos
17
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

55% (hard)

Question Stats:

74% (03:00) correct 26% (03:11) wrong based on 428 sessions

History

Date
Time
Result
Not Attempted Yet
If x and y are distinct positive integers, such that \((2^x)(16^{x−1})=2^{x^2}\) and \((3^y)(3^x)=3^{x^2−4x+5}\). What is the value of \((x)^{(y+1)}\) ?

A. 4
B. 8
C. 16
D. 32
E. 64
ShowHide Answer
Official Answer
C
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 21 Apr 2026
Posts: 109,724
Own Kudos:
810,386
 [2]
Given Kudos: 105,797
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,724
Kudos: 810,386
 [2]
If x and y are distinct positive integers, such that (2^x)(16^(x-1)) = 25 May 2017, 14:16
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
niteshwaghray
If x and y are distinct positive integers, such that \((2^x)(16^{x−1})=2^{x^2}\) and \((3^y)(3^x)=3^{x^2−4x+5}\). What is the value of \((x)^{(y+1)}\) ?

A. 4
B. 8
C. 16
D. 32
E. 64
Show more

Step 1:



\((2^x)(16^{x−1})=2^{x^2}\);

\((2^x)(2^{4(x−1)})=2^{x^2}\);

\(2^{x+4(x−1)})=2^{x^2}\);

\(x+4(x−1)=x^2\);

\(x=1\) or \(x=4\)

Step 2:



\((3^y)(3^x)=3^{x^2−4x+5}\);

\((3^{x+y}=3^{x^2−4x+5}\);

\(x+y=x^2−4x+5\);

\(x^2-5x+5=y\).

Now, if \(x = 1\), then \(y = 1\) too and we are told that x and y are distinct, thus \(x = 4\) and \(y = 1\).

Step 3:



Therefore, \((x)^{(y+1)}=4^{1+1}=16\).

Answer: C.
User avatar
MvArrow
User avatar
Joined: 06 Sep 2016
Last visit: 26 Jan 2019
Posts: 100
Own Kudos:
61
 [2]
Given Kudos: 100
Location: Italy
Schools: EDHEC (A$)
GMAT 1: 650 Q43 V37
GPA: 3.2
WE:General Management (Human Resources)
Schools: EDHEC (A$)
GMAT 1: 650 Q43 V37
Posts: 100
Kudos: 61
 [2]
If x and y are distinct positive integers, such that (2^x)(16^(x-1)) = 26 May 2017, 09:06
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
niteshwaghray
If x and y are distinct positive integers, such that \((2^x)(16^{x−1})=2^{x^2}\) and \((3^y)(3^x)=3^{x^2−4x+5}\). What is the value of \((x)^{(y+1)}\) ?

A. 4
B. 8
C. 16
D. 32
E. 64

Step 1:



\((2^x)(16^{x−1})=2^{x^2}\);

\((2^x)(2^{4(x−1)})=2^{x^2}\);

\(2^{x+4(x−1)})=2^{x^2}\);

\(x+4(x−1)=x^2\);

\(x=1\) or \(x=4\)

Step 2:



\((3^y)(3^x)=3^{x^2−4x+5}\);

\((3^{x+y}=3^{x^2−4x+5}\);

\(x+y=x^2−4x+5\);

\(x^2-5x+5=y\).

Now, if \(x = 1\), then \(y = 1\) too and we are told that x and y are distinct, thus \(x = 4\) and \(y = 1\).

Step 3:



Therefore, \((x)^{(y+1)}=4^{1+1}=16\).

Answer: C.
Show more

Is it correct in the first step to assume that X = 4 just noticing that 1 doesn't appear in the possible answers? Thanks!
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 21 Apr 2026
Posts: 109,724
Own Kudos:
810,386
Given Kudos: 105,797
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,724
Kudos: 810,386
If x and y are distinct positive integers, such that (2^x)(16^(x-1)) = 26 May 2017, 09:52
Kudos
Add Kudos
Bookmarks
Bookmark this Post
MvArrow
Bunuel
niteshwaghray
If x and y are distinct positive integers, such that \((2^x)(16^{x−1})=2^{x^2}\) and \((3^y)(3^x)=3^{x^2−4x+5}\). What is the value of \((x)^{(y+1)}\) ?

A. 4
B. 8
C. 16
D. 32
E. 64

Step 1:



\((2^x)(16^{x−1})=2^{x^2}\);

\((2^x)(2^{4(x−1)})=2^{x^2}\);

\(2^{x+4(x−1)})=2^{x^2}\);

\(x+4(x−1)=x^2\);

\(x=1\) or \(x=4\)

Step 2:



\((3^y)(3^x)=3^{x^2−4x+5}\);

\((3^{x+y}=3^{x^2−4x+5}\);

\(x+y=x^2−4x+5\);

\(x^2-5x+5=y\).

Now, if \(x = 1\), then \(y = 1\) too and we are told that x and y are distinct, thus \(x = 4\) and \(y = 1\).

Step 3:



Therefore, \((x)^{(y+1)}=4^{1+1}=16\).

Answer: C.

Is it correct in the first step to assume that X = 4 just noticing that 1 doesn't appear in the possible answers? Thanks!
Show more

Good observation. If x were 1, then the answer would be 1 because 1^(any number) = 1. So, yes we could say after Step 1, that x = 4.
User avatar
JeffTargetTestPrep
User avatar
Target Test Prep Representative
Joined: 04 Mar 2011
Last visit: 05 Jan 2024
Posts: 2,974
Own Kudos:
8,708
Given Kudos: 1,646
Status:Head GMAT Instructor
Affiliations: Target Test Prep
Expert
Expert reply
Posts: 2,974
Kudos: 8,708
If x and y are distinct positive integers, such that (2^x)(16^(x-1)) = 31 May 2017, 19:14
Kudos
Add Kudos
Bookmarks
Bookmark this Post
niteshwaghray
If x and y are distinct positive integers, such that \((2^x)(16^{x−1})=2^{x^2}\) and \((3^y)(3^x)=3^{x^2−4x+5}\). What is the value of \((x)^{(y+1)}\) ?

A. 4
B. 8
C. 16
D. 32
E. 64
Show more

Let’s simplify the first equation:

(2^x)(16^x-1) = 2^(x^2)

(2^x)(2^4)^(x-1) = 2^(x^2)

(2^x)(2^(4x-4)) = 2^(x^2)

2^(5x - 4) = 2^(x^2)

Because the bases are each 2, we can equate just the exponents:

5x - 4 = x^2

x^2 - 5x + 4 = 0

(x - 4)(x - 1) = 0

x = 4 or x = 1

Now let’s simplify the second equation:

(3^y)(3^x) = 3^(x^2 - 4x + 5)

3^(y + x) = 3^(x^2 - 4x + 5)

y + x = x^2 - 4x + 5

y = x^2 - 5x + 5

Recall from our earlier work that x^2 - 5x + 4 = 0; thus, y = x^2 - 5x + 5 = (x^2 - 5x + 4) + 1 = 0 + 1 = 1. Furthermore, recall that x = 4 or x = 1 and we are given that x and y are distinct integers, so x must be equal to 4. Thus, x^(y + 1) = 4^(1 + 1) = 4^2 = 16.

Answer: C
User avatar
pushpitkc
User avatar
Joined: 26 Feb 2016
Last visit: 19 Feb 2025
Posts: 2,800
Own Kudos:
6,235
Given Kudos: 47
Location: India
GPA: 3.12
Posts: 2,800
Kudos: 6,235
If x and y are distinct positive integers, such that (2^x)(16^(x-1)) = 01 Jun 2017, 02:33
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Since \((2^x)(16^{x−1})=2^{x^2}\)
We can clearly say that \(x+4x-4 = x^2\) because \(16 = 2^4\)
Therefore, \(x^2 = 5x-4\)

\((3^y)(3^x)=3^{x^2−4x+5}\)
Hence,\(y+x = {x^2−4x+5}\)
Substituting value of x^2 in this equation
y + x = 5x - 4 -4x + 5
y + x = x + 1 or y=1.

Substituting y=1, \((x)^{(y+1)}\) must be a perfect square.
From equation \(x^2 = 5x-4\) we can find out the value of x to be 1 or 4.
Only 4^2 is an option given in the answer options(Option C)
User avatar
jfranciscocuencag
User avatar
Joined: 12 Sep 2017
Last visit: 17 Aug 2024
Posts: 227
Own Kudos:
144
Given Kudos: 132
Posts: 227
Kudos: 144
If x and y are distinct positive integers, such that (2^x)(16^(x-1)) = 10 Feb 2019, 19:53
Kudos
Add Kudos
Bookmarks
Bookmark this Post
niteshwaghray
If x and y are distinct positive integers, such that \((2^x)(16^{x−1})=2^{x^2}\) and \((3^y)(3^x)=3^{x^2−4x+5}\). What is the value of \((x)^{(y+1)}\) ?

A. 4
B. 8
C. 16
D. 32
E. 64
Show more

Good night everyone!

Could someone please explain to me why \(2^{x^2}\) does not mean the following?

(2^m)^n = 2^mn

Isn't \(2^{x^2}\) the same as (2^x)^2?

Thank you so much in advance!8
User avatar
KanishkM
User avatar
Joined: 09 Mar 2018
Last visit: 18 Dec 2021
Posts: 756
Own Kudos:
512
 [1]
Given Kudos: 123
Location: India
Schools: DeGroote'21 Desautels '21 NUS '21
Schools: DeGroote'21 Desautels '21 NUS '21
Posts: 756
Kudos: 512
 [1]
If x and y are distinct positive integers, such that (2^x)(16^(x-1)) = 10 Feb 2019, 20:09
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
jfranciscocuencag
niteshwaghray
If x and y are distinct positive integers, such that \((2^x)(16^{x−1})=2^{x^2}\) and \((3^y)(3^x)=3^{x^2−4x+5}\). What is the value of \((x)^{(y+1)}\) ?

A. 4
B. 8
C. 16
D. 32
E. 64

Good night everyone!

Could someone please explain to me why \(2^{x^2}\) does not mean the following?

(2^m)^n = 2^mn

Isn't \(2^{x^2}\) the same as (2^x)^2?

Thank you so much in advance!8
Show more

Hey jfranciscocuencag

(2^m)^n = 2^mn , The presence of bracket makes all the difference, As per PEMDAS, we solve the parenthesis first

But if you see here, there is no parenthesis, because of that i will first solve the exponent and not the expression \(2^{x^2}\)

Have a look here as well
https://www.mathsisfun.com/operation-order-pemdas.html
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 21 Apr 2026
Posts: 109,724
Own Kudos:
810,386
Given Kudos: 105,797
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,724
Kudos: 810,386
If x and y are distinct positive integers, such that (2^x)(16^(x-1)) = 10 Feb 2019, 21:55
Kudos
Add Kudos
Bookmarks
Bookmark this Post
jfranciscocuencag
niteshwaghray
If x and y are distinct positive integers, such that \((2^x)(16^{x−1})=2^{x^2}\) and \((3^y)(3^x)=3^{x^2−4x+5}\). What is the value of \((x)^{(y+1)}\) ?

A. 4
B. 8
C. 16
D. 32
E. 64

Good night everyone!

Could someone please explain to me why \(2^{x^2}\) does not mean the following?

(2^m)^n = 2^mn

Isn't \(2^{x^2}\) the same as (2^x)^2?

Thank you so much in advance!8
Show more

\(a^m^n=a^{(m^n)}\) and not \((a^m)^n\) (if exponentiation is indicated by stacked symbols, the rule is to work from the top down). Thus, \(2^{x^2}=2^{(x^2)}\) and not \((2^x)^2=2^{2x}\)

On the other hand, \((a^m)^n=a^{mn}\).

8. Exponents and Roots of Numbers



Check below for more:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread
User avatar
luisdicampo
User avatar
Joined: 10 Feb 2025
Last visit: 19 Apr 2026
Posts: 480
Own Kudos:
73
Given Kudos: 328
Products:
My Reviews Forum Quiz
Posts: 480
Kudos: 73
If x and y are distinct positive integers, such that (2^x)(16^(x-1)) = 10 Mar 2026, 07:44
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Deconstructing the Question

Interpret the equations as

\((2^x)(16^{x-1})=2^{x^2}\)

and

\((3^y)(3^x)=3^{x^2-4x+5}\)

We need to find

\(x(y+1)\)

The key idea is that when the bases are the same, we can equate exponents.

Step-by-step

Start with

\((2^x)(16^{x-1})=2^{x^2}\)

Since

\(16=2^4\)

we get

\(16^{x-1}=(2^4)^{x-1}=2^{4x-4}\)

So the left side becomes

\(2^x \cdot 2^{4x-4}=2^{5x-4}\)

Thus

\(2^{5x-4}=2^{x^2}\)

So

\(x^2=5x-4\)

\(x^2-5x+4=0\)

\((x-1)(x-4)=0\)

So

\(x=1\)

or

\(x=4\)

Now use the second equation

\((3^y)(3^x)=3^{x^2-4x+5}\)

The left side is

\(3^{x+y}\)

So

\(x+y=x^2-4x+5\)

\(y=x^2-5x+5\)

Test \(x=1\)

\(y=1-5+5=1\)

But \(x\) and \(y\) must be distinct, so this case is invalid.

Test \(x=4\)

\(y=16-20+5=1\)

This works.

Now compute

\(x(y+1)=4(1+1)=8\)

Answer B: 8
Moderators:
Bunuel
Math Expert
109724 posts
Krunaal
Tuck School Moderator
853 posts