Last visit was: 14 Jul 2024, 07:40 It is currently 14 Jul 2024, 07:40
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
SORT BY:
Date
Tags:
Show Tags
Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 94342
Own Kudos [?]: 640711 [25]
Given Kudos: 85011
Send PM
Most Helpful Reply
GMAT Club Legend
GMAT Club Legend
Joined: 03 Jun 2019
Posts: 5275
Own Kudos [?]: 4178 [7]
Given Kudos: 160
Location: India
GMAT 1: 690 Q50 V34
WE:Engineering (Transportation)
Send PM
Intern
Intern
Joined: 03 Aug 2009
Posts: 26
Own Kudos [?]: 68 [5]
Given Kudos: 27
Send PM
Current Student
Joined: 16 Jan 2019
Posts: 623
Own Kudos [?]: 1477 [2]
Given Kudos: 142
Location: India
Concentration: General Management
GMAT 1: 740 Q50 V40
WE:Sales (Other)
Send PM
Re: If x^(a-b) = 2 and x^(a+b) = 32, and if x, a, and b are greater than [#permalink]
1
Kudos
1
Bookmarks
\(x^{(a−b)}=2\) and \(x^{(a+b)}=32\)

\(\frac{x^a}{x^b} = 2\) and \(x^a*x^b=32\)

Solving the two equations we get \(x^a = 8\) and \(x^b=4\) - II must be true

Since we don't know that \(x, a\) or \(b\) are integers we cannot assume that \(a=3\) - I need not be true

Substitute \(b=\frac{2}{3}a\) or \(a=\frac{3}{2}b\) in \(x^a = 8\)

\(x^{3b/2}=2^3\) or \(x^{b/2}=2\) or

\(x^b=4\) - TRUE - III must be true

So II and III must be true

Answer is (D)
General Discussion
Manager
Manager
Joined: 27 May 2010
Posts: 105
Own Kudos [?]: 262 [2]
Given Kudos: 22
Send PM
If x^(a-b) = 2 and x^(a+b) = 32, and if x, a, and b are greater than [#permalink]
2
Kudos
x^(a−b)=2

x^(a+b)=32=2^5 = (x^(a-b))^5

x^(a+b) = (x^5(a-b))
a+b=5(a-b)
4a=6b
Simplifying gives b = 2a/3, III is correct.

Substituting in x^(a−b)=2
x^(3b/2 - b) = 2
x^(b/2) = 2
x^b = 4, II is correct

Option D.

Posted from my mobile device

Originally posted by prashanths on 15 Jul 2019, 08:37.
Last edited by prashanths on 15 Jul 2019, 23:29, edited 1 time in total.
Senior Manager
Senior Manager
Joined: 06 Jun 2019
Posts: 313
Own Kudos [?]: 974 [2]
Given Kudos: 655
Location: Uzbekistan
Send PM
If x^(a-b) = 2 and x^(a+b) = 32, and if x, a, and b are greater than [#permalink]
1
Kudos
1
Bookmarks
This question can be a good trap if we assume that \(2\) is 1st power and \(32\) is the 5th power of \(2\). The test writer wants us to fall into this trap. If we hastily equate \((a-b)\) to \(1\) and \((a+b)\) to \(5\), then we will indeed find ourselves there. Being suspicious might keep us from such precipitate assumptions.

FIRST STEP: What we know for sure is \(x^{(a−b)}=2\) and \(x^{(a+b)}=32\). We have three unknowns \(x\), \(a\), and \(b\). We should also note that \(x^{(a+b)}=32\) is the 5th power of \(x^{(a−b)}=2\). So we can increase \(x^{(a−b)}=2\) to its 5th power and subtract it from the other:

So if \((x^{(a−b)})^5=2^5\), then we have \(x^{(5a-5b)}=32\). Next, by subtracting we have \(x^{(5a-5b)}=x^{(a+b)}\). Since x>0, the powers must be equal. \(5a-5b=a+b\), so we have \(b=\frac{2}{3}*a\). Bingo! So III must be true!

SECOND STEP: Can we figure out anything else with the help of III? If \(b=\frac{2}{3}*a\) and \(x^{(a−b)}=2\), then by substituting \(a\) with \(b\) we might prove II too, let's check (that's the way we need to think during the exam). If \(a=\frac{3}{2}*b\), then by substitution we have \(x^{(1.5b - b)}=2\). If \(x^{0.5b}=2\), then \(x^b=4\). Bingo! So II also must be true!

THIRD STEP: Considering II and III, can we now conclude that I also must be true or \(a=3\)? Let's use what we already have. If \(x^b=4\) and \(x^{(a+b)}=32\) or \(x^a*x^b=32\), then after substitution we have \(x^a*4=32\) or \(x^a=8\).

If \(x^a=8\), then a=3? We need to prove that \(a\) is \(3\) or instead \(a\) can be any number. So far we know that \(x^a=8\), \(x^b=4\), and \(\frac{a}{b}=\frac{3}{2}\). Let's try some numbers:

1. If \(a=3\), then \(x^a=8\) or \(x^3=8\) or \(x=2\).
If \(b=2\), then \(x^b=4\) or \(x^2=4\) or \(x=2\). We can see that \(a=3\) when \(x=2\).

2. If \(a=6\), then \(x^a=8\) or \(x^6=8\) or \(x=\sqrt{2}\).
If \(b=4\), then \(x^b=4\) or \(x^4=4\) or \(x=\sqrt{2}\). We can see that \(a=6\) when \(x=\sqrt{2}\).

3. If \(a=90\), then \(x^a=8\) or \(x^{90}=8\) or \(x=\sqrt[30]{2}\).
If \(b=60\), then \(x^b=4\) or \(x^{60}=4\) or \(x=\sqrt[30]{2}\). We can see that \(a=90\) when \(x=\sqrt[30]{2}\)

Conclusion: depending on \(x\), \(a\) can be any positive number, not only \(3\). Thus I can be true, but not must be true.

Hence D

Originally posted by JonShukhrat on 15 Jul 2019, 09:03.
Last edited by JonShukhrat on 15 Jul 2019, 21:37, edited 1 time in total.
Retired Moderator
Joined: 19 Oct 2018
Posts: 1868
Own Kudos [?]: 6632 [1]
Given Kudos: 705
Location: India
Send PM
Re: If x^(a-b) = 2 and x^(a+b) = 32, and if x, a, and b are greater than [#permalink]
1
Kudos
It's a must be true question. Hence if any statement is not true for any positive real values of x, a and b
\(x^{a-b}= 2\)
\(a-b= log_x2\)......(1)

\(x^{a+b}=32\)
\(a+b= log_x32\)= \(log_x2^5\)= \(5log_x2\)......(2)

Add equation (1) and (2)
2a= \(6log_x2\)
a=\(3log_x2\)......(3)


Subtract equation (1) from (2)
2b= \(4log_x2\)
b=\(2log_x2\).....(4)


I) a=\(3log_x2\)
We can see that value of a is dependent on x.
If x=2, a=3.
If \(x=\sqrt{2}\), a=6.

Must not true

II
From equation 4
b=\(2log_x2\)
b=\(log_x2^2\)
b=\(log_x4\)
\(x^b=4\)

Must be true

III
Divide equation 4 by equation 3
\(\frac{b}{a}\)= \(\frac{2log_x2}{3log_x2}\)
\(\frac{b}{a}\)=\(\frac{2}{3}\)
\(b=(\frac{2}{3})*a\)

Must be true

Option D
SC Moderator
Joined: 25 Sep 2018
Posts: 1100
Own Kudos [?]: 2236 [2]
Given Kudos: 1665
Location: United States (CA)
Concentration: Finance, Strategy
GPA: 3.97
WE:Investment Banking (Investment Banking)
Send PM
Re: If x^(a-b) = 2 and x^(a+b) = 32, and if x, a, and b are greater than [#permalink]
2
Kudos
Solution:

Lets begin with solving the equation ,

Our focus first should be getting the same bases.

i)\(x^{(a-b)}\) =2 ,
ii)\(x^{(a+b)}\) = \(2^5\)

Raising power 5 to both the sides of the equation one, we get \(x^{5(a-b)}\) = \(2^5\)

Equating the bases in i) & ii)

We get,



5a - 5b = a + b,
Therefore 6b =4a
b = \(\frac{2}{3}\)a
Hence condition III is satisfied here.

Substituting value of b in equation i,

we get x^\({(a- b)}\) = 2,

\(x^{(\frac{3}{2b} - b)}\) = 2,
therefore, \(x^{(\frac{b}{2})}\) = 2,
If we square both the sides, we get ,

\(x^b\) = 4 which satisfies condition II.

Now, we know from the question stem that \(x^{(a-b)}\)= 2,

We also know that condition III is being satisfied, hence we can take b=\(\frac{2}{3}\)a and substitute it in above equation.

We get that, \(x^{(a-\frac{2}{3}a)}\) solving this, we get \(x^\frac{a}{3}\)= 2, squaring both the sides we get \(x^a\) = \(2^3\)
If x= 2 , then x =3 since the bases will be equal, However , if x is 64 then a will be \(\frac{1}{2}\). we are getting different values for a so this condition cannot be satisfied

Hence the answer is D
Ross School Moderator
Joined: 07 Dec 2018
Posts: 101
Own Kudos [?]: 108 [2]
Given Kudos: 111
Location: India
Send PM
Re: If x^(a-b) = 2 and x^(a+b) = 32, and if x, a, and b are greater than [#permalink]
1
Kudos
1
Bookmarks
I. a=3
II. \(x^b=4\)
III. b=\(\frac{2}{3}\)a

\(x^{a−b}\)=2
\(x^{a+b}\)=32

Looking at answer choices, we should try to find possible values of a, b & x.

\(x^{a+b}\) / \(x^{a−b}\) = 16
\(x^{a+b-a+b}\)=16
\(x^{2b}\)=16

\(x^{a+b}\) * \(x^{a−b}\) = 64
\(x^{a+b+a-b}\)=64
\(x^{2a}\)=64

Statement 1 : a=3

\(x^{2a}\)=64
As per this equation if a=3 => x=3
But, a can also be 1. In that case x=8
So, can be true but NOT must be true.

Statement 2 : \(x^b=4\)

\(x^{2b}\)=16
If x=4 then b=1 =>\(x^b=4\)
If x=16 then b=1/2 =>\(x^b=4\)
If x=2 then b=2 =>\(x^b=4\)
Hence, Statement 2 is a MUST BE TRUE statement.

Statement 3 : b=\(\frac{2}{3}\)a

We have already figured following out in Statement 2 that,
If x=4 then b=1
If x=16 then b=1/2
If x=2 then b=2

If we place these values of x in \(x^{2a}\)=64
We get,
If x=4 then b=1 => x=3/2
If x=2 then b=2 => x=3
Which proves relation between a & b is indeed b=\(\frac{2}{3}\)a

Hence, statement 3 is also a MUST BE TRUE statement.


Ans should be (D)
User avatar
Non-Human User
Joined: 09 Sep 2013
Posts: 33968
Own Kudos [?]: 851 [0]
Given Kudos: 0
Send PM
Re: If x^(a-b) = 2 and x^(a+b) = 32, and if x, a, and b are greater than [#permalink]
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.
GMAT Club Bot
Re: If x^(a-b) = 2 and x^(a+b) = 32, and if x, a, and b are greater than [#permalink]
Moderator:
Math Expert
94342 posts