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

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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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New post 15 Jul 2019, 18:56
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
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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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New post 15 Jul 2019, 20:24
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
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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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New post 15 Jul 2019, 20:26
Every value has to be true.
E is the right answer.
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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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New post 15 Jul 2019, 21:08
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.
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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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New post 15 Jul 2019, 21:14
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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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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New post 15 Jul 2019, 21:22
\(x^{a-b}\)=\(2^{1}\)
\(x^{a+b}\)=\(2^{5}\)
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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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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New post 15 Jul 2019, 21:51
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
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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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New post 15 Jul 2019, 22:04
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)
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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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New post 15 Jul 2019, 23:56
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
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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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New post 16 Jul 2019, 00:00
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
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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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New post 16 Jul 2019, 00:34
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

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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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New post 16 Jul 2019, 01:08
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
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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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New post 16 Jul 2019, 01:32
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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If x^(a-b) = 2 and x^(a+b) = 32, and if x, a, and b are greater than  [#permalink]

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New post 16 Jul 2019, 02:09
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
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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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New post 16 Jul 2019, 03:33
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
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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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New post 16 Jul 2019, 04:10
B. II.

II must be true. Subsequently, I & III would follow as true.
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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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New post 16 Jul 2019, 04:18
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
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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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New post 16 Jul 2019, 05:30
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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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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New post 16 Jul 2019, 05:34
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).
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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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New post 16 Jul 2019, 05:57
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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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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