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Manager  S
Joined: 17 Jul 2014
Posts: 111
Re: If x^(a-b) = 2 and x^(a+b) = 32, and if x, a, and b are greater than  [#permalink]

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Given
$$x^{(a−b)}=2^1$$
$$x^{(a+b)}=32$$ = $$2^{5}$$
x, a, and b are > 0, which of the following must be true?

Derive from above
a-b = 1
a+b = 5
a = 3, b = 2

I. a=3 -> this is exactly as given, so true

II. $$x^b$$=4 = 2^2 ; b = 2 - as given, so true

III. b= $$\frac{2}{3}$$*a = $$\frac{2}{3}$$ * 3 = 2 - as given, so true

E. I, II, and III - is the answer
Director  D
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Re: If x^(a-b) = 2 and x^(a+b) = 32, and if x, a, and b are greater than  [#permalink]

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If $$x^{(a−b)}=2$$ and $$x^{(a+b)}=32$$, and if x, a, and b are greater than 0, which of the following must be true?

I. a=3
II. $$x^b=4$$
III. $$b=(2/3)a$$

A. I only
B. II only
C. I and II
D. II and III
E. I, II, and III

given $$x^{(a-b)}$$ = 2
and $$x^{(a+b)}$$ = 32
multiply both equations = $$x^{2a}$$= $$2^6$$
thus $$x^a$$= $$2^3$$

but this can also be written as 8^1
thus a can be 3 or 1
hence 1 must not be true

divide both given equation
$$x^{2b}$$ = $$16$$= $$2^4$$

or $$x^b =4$$
hence 2 must be true

now
with a = 3
and b =2
equation 3 satisfies
but with a as 1 and b = 2 this does not satisfy
hence again 3 must not be true

Thus only 2 is possible
Answer is B
Manager  S
Joined: 10 Aug 2018
Posts: 220
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Concentration: Strategy, Operations
WE: Operations (Energy and Utilities)
Re: If x^(a-b) = 2 and x^(a+b) = 32, and if x, a, and b are greater than  [#permalink]

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Every value has to be true.
E is the right answer.
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Manager  S
Joined: 27 Mar 2018
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Location: India
Re: If x^(a-b) = 2 and x^(a+b) = 32, and if x, a, and b are greater than  [#permalink]

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1. a=3
x^(a-b) * x^(a+b) = 2*32
=> x^2a = 64 or x^6=64 then x=2 (greater than 0)
$$\frac{x^(a+b)}{x^(a-b)}$$ = 16
=>x^2b = 16 or x^b=4
if x=2, then b=2(greater than 0). True

2. x^(a+b)/x^(a-b) = 16
=>x^2b = 16 or x^b=4. True

3. b=(2/3)a
x^$$\frac{a}{3}$$ =2 or x^a = 8 => x^b=4
x^$$\frac{5a}{3}$$=32 or x^a=8
x, a and b have multiple values that can be true.

Hence IMO option C.
_________________
Thank you for the kudos. You are awesome! Manager  S
Joined: 29 May 2019
Posts: 99
Re: If x^(a-b) = 2 and x^(a+b) = 32, and if x, a, and b are greater than  [#permalink]

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If x^(a−b)=2 and x^(a+b)=32, and if x, a, and b are greater than 0, which of the following must be true?

x^(a−b)=2
We can say that a-b is positive as we do not get answer in fraction. So a > b.
And x^anything = 2. We can consider anything as 1. No other value fits.
So x is 2 and a-b = 1

x^(a+b)=32
if x is 2 then a + b = 5

if a + b = 5 and a-b = 1 the we can say a = 3 and b = 2

I. a=3 : We have in the above explanation
II. x^b=4 : We have in the above explanation
III. b=(2/3)a: We can say this from the above explanation.

Answer: E

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Joined: 12 Jan 2018
Posts: 112
Re: If x^(a-b) = 2 and x^(a+b) = 32, and if x, a, and b are greater than  [#permalink]

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$$x^{a-b}$$=$$2^{1}$$
$$x^{a+b}$$=$$2^{5}$$
a+b=5, a-b=1 => a=3, b=2

I. a=3, True
II.$$x^{b}$$=4, b=2, x=2,$$2^{2}$$=4,True
III.b=($$\frac{2}{3}$$) a , b=($$\frac{2}{3}$$)*3=> b=2, True

Ans. E

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Manager  S
Joined: 24 Jan 2019
Posts: 103
Location: India
Concentration: Strategy, Finance
GPA: 4
Re: If x^(a-b) = 2 and x^(a+b) = 32, and if x, a, and b are greater than  [#permalink]

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1
If x^(a−b)=2 and x^(a+b)=32, and if x, a, and b are greater than 0, which of the following must be true?

I. a=3
[x,a,b] can be [2, 3, 2], [sqrt(2), 6, 4] and so on. So, value of a,b depends on the value of X............eliminated.

II. x^b=4
x^(a+b)=16*2 = 16*(x^(a-b)).........simplify this and we will get X^b = 4..............correct.

III. b=(2/3)a
X^(a+b) = 2^5 = (x^(a-b))^5.........simplify this and we will get b = (2/3)a...........correct.

Answer : D
Manager  S
Joined: 07 Dec 2018
Posts: 111
Location: India
Concentration: Technology, Finance
GMAT 1: 670 Q49 V32 Re: If x^(a-b) = 2 and x^(a+b) = 32, and if x, a, and b are greater than  [#permalink]

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1
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)
Intern  B
Joined: 11 Jun 2014
Posts: 23
Re: If x^(a-b) = 2 and x^(a+b) = 32, and if x, a, and b are greater than  [#permalink]

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Answer- B

Exp: x^(a+b)/x^(a-b)=16
x^2b=16
x^b=4 (ii)correct
if b=1 & x=4 that will satisfy ii
Again x^a2-b2=8^2
x=4,b=1,x^a2-1=4^3
a^2-1=3
a=2 hence not satisfy i & iii
Senior Manager  P
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Re: If x^(a-b) = 2 and x^(a+b) = 32, and if x, a, and b are greater than  [#permalink]

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If x^(a−b)=2 and x^(a+b)=32, and if x, a, and b are greater than 0, which of the following must be true?

I. a=3
II. x^b=4
III. b=2/3 a

given x^(a-b) = 2^1 and
x^(a+b ) =32

=>$$[fraction]x^a*x^b[/fraction]$$ =2^5

=> x = 2 , a+b = 5 , a-b = 1
==> x =2 , a = 3 , b =2

I . a = 3 ---True
II . X^b =4 , ---2^2 = 4 --True
III. b = 2/3 *a =>2/3 * 3 = 2 =b --True

E. I, II, and III----Correct
Manager  S
Joined: 10 Nov 2014
Posts: 59
Location: United States
Concentration: Marketing, Strategy
Schools: Goizueta '22, IIM
Re: If x^(a-b) = 2 and x^(a+b) = 32, and if x, a, and b are greater than  [#permalink]

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Soln:

X^a-b=2 and x^ a+b=32

2^1=1 and 2^5=32

Therefore a-b=1 and a+b=5, solving them for value of a and b , we get a=3 and b=2

Lets check each statement
I ) a=3 TRUE
II) x^b=4, x=2 and b=2 so 2^2 = 4 TRUE
III) b=2/3* a b= 2 and a=3 TRUE

Ans Choice E

Dream it ! Achieve it!
Manager  G
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Re: If x^(a-b) = 2 and x^(a+b) = 32, and if x, a, and b are greater than  [#permalink]

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1
Given,

1. x^(a-b) = 2
2. x^(a+b) = 32
3. x > 0, a > 0, b > 0

Note that it is not mentioned that x, a, b are integers.
So they can be fractions too.

We have 2 cases here now.

Case 1: x = 2

1. x^(a-b) = 2
=> a - b = 1 .... Equation 1

2. x^(a+b) = 32 = 2^5
=> a + b = 5 .... Equation 2

Solving for equations 1 and 2 we get,
a = 3, b = 2

Case 2: x = 32

1. x^(a-b) = 2
=> a - b = 32^(1/5) .... Equation 3

2. x^(a+b) = 32
=> a + b = 1 .... Equation 4

Solving for equations 3 and 4 we get,
a = 3/5, b = 2/5

Now let us consider the options given after we have found the 2 set of values of a and b.

I. a = 3
This is true in case 1 and not in case 2.

Hence not always true.

II. x^b = 4
This is true in case 1 and case 2.

Hence always true.

III. b = (2/3)*a
This is true in case 1 and case 2.

Hence always true.

Answer: D
Director  P
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Re: If x^(a-b) = 2 and x^(a+b) = 32, and if x, a, and b are greater than  [#permalink]

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1
$$\frac{x^{a+b}}{x^{a-b}}$$ = 32/2
--> $$x^{2b}$$ = 16
--> $$x^b = 4$$ (II must be true)

If $$x^b = 4$$, Possible cases
Case 1: $$4^1 = 4$$
--> x = 4, b = 1.
--> $$4^{a - 1} = 2 = 2^1$$
--> $$2^{2a - 2} = 2^1$$
--> 2a - 2 = 1
--> a = 3/2
--> (x, a, b) = (4, 3/2, 1)

I. a = 3 --> Not True
III. b = 2/3a = 2/3*3/2 = 1 --> True

Case 2: $$2^2 = 4$$
--> x = 2, b = 2
--> $$2^{a - 2} = 2 = 2^1$$
--> a - 2 = 1
--> a = 3
--> (x, a, b) = (2, 3, 2)

I. a = 3 --> True
III. b = $$\frac{2}{3}$$a = $$\frac{2}{3}$$*3 = 2 --> True

So, II & III are always True

IMO Option D

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Manager  G
Joined: 31 May 2018
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Concentration: Finance, Marketing
If x^(a-b) = 2 and x^(a+b) = 32, and if x, a, and b are greater than  [#permalink]

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1
x^(a-b)=2 --(1) and x^(a+b)=32--(2)

x^(a+b)=2^5

x^(a+b)=x^5(a-b)-----(from equation-1)

a+b = 5(a-b)
6b = 4a
b = $$\frac{2a}{3}$$

I. a=3
b = $$\frac{2a}{3}$$
when b=2 then a=3
b=4 then a=6

a=3 could be true but not always true
since this is must be true question a=3 should always be true
but this is not the case here so (OUT)

II. $$X^b$$=4

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

x^($$\frac{3b}{2}$$-b) = 2 and x^($$\frac{3b}{2}$$+b) = $$2^5$$

x^($$\frac{b}{2}$$) = 2 and x^$$\frac{5b}{2}$$ = $$2^5$$

x^b = 4 (squaring both sides) and x^b = 4 (raising both sides to the power$$\frac{2}{5}$$)
this is always true

III. b = $$\frac{2a}{3}$$
solving equation (1)&(2)
we get

b = $$\frac{2a}{3}$$
so this is always true

D is the answer
Intern  B
Joined: 08 Jul 2019
Posts: 37
Re: If x^(a-b) = 2 and x^(a+b) = 32, and if x, a, and b are greater than  [#permalink]

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x^(a-b)=2^1, we can deduce that x is 2 and a-b=1
x^(a+b)=2^5, we can deduce that x is 2, and a+b=5,
a-b=1
a+b=5 (add both)
2a=6
a=3, b=2 (I is true)
x^b=2^2, 2^b=2^2, from above we know that b is 2, this option also says b=2 (II is true)
b=2a/3, substitute with numbers from I. 2=2*3/3=2 (III is true)
Answer should be E
Intern  B
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Schools: ISB '21
GMAT 1: 600 Q40 V34 GPA: 4
Re: If x^(a-b) = 2 and x^(a+b) = 32, and if x, a, and b are greater than  [#permalink]

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B. II.

II must be true. Subsequently, I & III would follow as true.
Manager  G
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Re: If x^(a-b) = 2 and x^(a+b) = 32, and if x, a, and b are greater than  [#permalink]

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1
Okay, let's get down to solving this:

x^(a-b) = 2^1 and x^(a+b) = 2^5 (i.e. 32 (given))

Considering that x = 2 is an easy trap to fall for. HOWEVER, x can have any number of values, let's say x = 2^(1/2) or 2^(1/4), in those cases, a-b would be 2 and 4 respectively and a+b would be 10 and 20 respectively.

Looking at this, we can see that it's not a "must be true" condition that a = 3. Yes, a CAN be 3, but a can ALSO have an infinite number of other values as well. So (I) is not true.

x^b = 4, yes, this will always be true for all values of x and b. If x = 2 (a = 3 and b = 2), and hence x^b = 2^2 = 4

Similarly, if x = 2^(1/4) (a = 12 and b = 8), and hence x^b = (2^(1/4))^(8) = 2^2 = 4

You could try out more (if time permits), but if you look at it logically, you won't need to and this would help you understand that this stays true for all sets of values of x and b. Hence, x^b = 4 is always true. (II) is true.

Looking at what we've derived above, we can see that when a = 12, b = 8; when a = 6, b = 4; when a = 3, b = 2, and within no time, we can conclude that b = 2/3(a), and hence (III) is also true.

Thus, the correct answer choice is (D) II and III
Intern  B
Joined: 10 Aug 2017
Posts: 29
Re: If x^(a-b) = 2 and x^(a+b) = 32, and if x, a, and b are greater than  [#permalink]

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If x^(a−b)=2 and x^(a+b)=32, and if x, a, and b are greater than 0, which of the following must be true?

I. a=3 --> Not MBT
II. xb=4 --> MBT
III. b=(2/3)a --> MBT

A. I only
B. II only
C. I and II
D. II and III
E. I, II, and III

Answer is D

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Intern  S
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Re: If x^(a-b) = 2 and x^(a+b) = 32, and if x, a, and b are greater than  [#permalink]

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The only solutions for these equations are $$x^{a−b}=2^1$$ and $$x^{a+b}=2^5$$.

Hence, x=2.
a-b=1 and a+b=5.
Solving, we get a=3 and b=2.

We find that I, II and III are correct with these values of x,a and b.

Hence, option(E).
Intern  B
Joined: 02 Jun 2013
Posts: 43
Re: If x^(a-b) = 2 and x^(a+b) = 32, and if x, a, and b are greater than  [#permalink]

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If x^(a−b)=2 and x^(a+b)=32, and if x, a, and b are greater than 0,

Given ,

x^(a-b)= 2
so here x= 2 --- equation 1
and a-b= 1 ----equation 2

x^(a+b) = 32
since we have x=2
so a+b= 5 ----equation 3

With equation 2 and 3 -

a= 3 , b=2

now we can try all given scenario

1. a=3 - possible
2. x^(b)= 2^2= 4 - possible
3. b= (2/3)a - possible

So answer is E
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Nothing comes easy ! Neither do i want !! Re: If x^(a-b) = 2 and x^(a+b) = 32, and if x, a, and b are greater than   [#permalink] 16 Jul 2019, 05:57

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