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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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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, 08:00
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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. \(b = (\frac{2}{3})a\)

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


 

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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 Updated on: 16 Jul 2019, 06:28
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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

\(x^(a-b) =2\) (1)
\(x^(a+b)=32\) (2)
Multiplying both equations
\(x^2a = 64=2^6=4^3\)
\(x^a=8\)
Dividing (2) by (1)
\(x^2b=16 =2^4=4^2\)
\(x^b=4\)

I. a=3
\(x^a=8 if x=2=>a=3 but if x=4=> a=\frac{3}{2}\)
COULD BE TRUE BUT NOT MUST BE TRUE

\(II. x^b=4\)
\(x^b=4\)
MUST BE TRUE

III. b=(\frac{2}{3})a
\(x^b=4=8^\frac{2}{3}=x^\frac{2a}{3}\)
\(b=\frac{2}{3}a\) if x<>0 and x<>1 which are true. since \(x>0 and x^b=4\)
MUST BE TRUE

IMO D
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Originally posted by Kinshook on 15 Jul 2019, 08:28.
Last edited by Kinshook on 16 Jul 2019, 06:28, edited 8 times in total.
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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, 08:28
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\(x^{(a−b)}=2\) 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)
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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, 08:12
1. x^(a-b)=2^1 -> a-b=1
2. x^(a+b)=32=2^5 -> a+b=5

Resolving 1 & 2:
a=3,b=2

which satisfies all the cases
Hence 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, 08:13
If x(a−b)=2and 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. xb=4
III. b=(2/3)a

This is a fairly simple question. Put each of the answer choices in the statement above. all the choices are must be true.


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

Hence 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]

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New post 15 Jul 2019, 08:15
1
solving, 5(a-b)=(a+b)
gives a= 3b/2
using this x^b=4
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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 Updated on: 15 Jul 2019, 09:52
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1
first of all, be careful, cause nowhere it says that X, a and b ARE INTEGERS. This is important, cause one could assume that since \(x^{(a−b)}=2\)
then x = 2, and \((a-b) =1\). As a valid option X could be \(\sqrt{2}\) and \((a-b) =2\), therefore 1st stament is NOT always true.

So, without knowing X, we may work with powers:
\([x^{(a+b)}=32] => [x^{(a+b)}=2^{5}]\), since X>0 \(=> [(a-b) = (a+b)/5] => [(a+b) = 5(a-b)] => [6b=4a] =>[b = (\frac{2}{3})*a]\), which is 3rd statement.

Now, lets assume \(a-b = y\), then \(a+b = 5y\) (see above) \(=> [a-b - (a+b) = y-5y] => [-2b= -4y] => [b=2y]\). We know that \(x^{y}= 2\), powering both sides by 2 we get \(x^{2y} = 4\), and since \(2y = b\), therefore \(x^{b}=4\), which is 2nd statement.

Therefore, 2 and 3 must be true and answer is D

Originally posted by berdibekov on 15 Jul 2019, 08:19.
Last edited by berdibekov on 15 Jul 2019, 09:52, edited 2 times in total.
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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, 08:22
Answer is E. I, 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 15 Jul 2019, 08:23
x^a/x^b = 2

x^a = 2 x^b......(1)

x^a * x^b = 32

By equating eq (1)

x^2b = 2^4
x^b = 2^2 = 4

and a = 3

This satisfies I and II

3b = 2a
3*2 = 2*3
6 = 6

Hence 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 15 Jul 2019, 08:24
If x(a−b)=2x(a−b)=2 and x(a+b)=32x(a+b)=32, and if x, a, and b are greater than 0, which of the following must be true?

so
x^a-b=2 and x^a+b= 32
from given info we can determine that
x=2 , a=3 and b=2
so
I. a=3 ; TRUE
II. x^b=4 ;TRUE
III. b=(2/3)* a ; TRUE
IMO 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, 08:25
IMO answer is E

multiplying x^(a−b)=2 and x^(a+b)=32, we get x^a = 8 and dividing both we get x^b = 4

x^b = 4 satisfies II
equating powers
a+b = 5a-5b => a = 3b/2

values which satisfy these set of conditions would be x= 2, a= 3 and b =2,
not sure if there are any decimals possible but I don't think so.

so 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, 08:28
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

\(x^(a−b)=2\)
This implies . \(x^(a−b)=2^1\)
So, \(a -b = 1\).................................1

\(x^(a+b)=32\)
This implies . \(x^(a−b)=2^5\)
So, \(a +b = 5\)...............................2

And x = 2

Solving equation 1 & 2 - a =3 & b= 2.

Substitute respective values.

1. TRUE
2. TRUE
3. 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, 08:36
If x(a−b)=2x(a−b)=2 and x(a+b)=32x(a+b)=32, and if x, a, and b are greater than 0, which of the following must be true?

I. a=3a=3
II. xb=4xb=4
III. b=(23)ab=(23)a

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

Combining the two equation of the question x^(a-b) = 2 and X^(a+b) = 32 we can write x^(a+b) = x^(5*(a-b))
simplifying the two sides we get a+b = 5 (a-b) which will give b = (2/3)a hence option 3 is correct. Now lets look at the options in answer. We have either 1,2 or 3 or 2 or 3.
So lets evaluate 1.
For x^(a-b) = 2 and X^(a+b) = 32 to be true both a and b has to be integer.
Now since a=3/2*b we will have b=2 and a=3

hence answer = E
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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 Updated on: 15 Jul 2019, 23:29
2
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.

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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.
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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, 08:38
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x^(a−b)=2 and x^(a+b)=32 means X=2, a = 3 and b=2 or X=root2, a=6 and b=4 and similar cases.

I. a=3 - Need not be true as A can be 6
II. x^b=4 - Must be true 2^2 or root2^4
III. b=(2/3)a - Must be true (fits both cases and similar iterations)

IMO D - II and III only
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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, 08:39
1
D

From the given equations: x^(a−b)=2, and x^(a+b)=32 = 2^5. Comparing both equations, we get 5*(a-b) = a+b or 2a=3b. Note that a = 3 is a possible solution but doesn't have to be, a can be any multiple of 3. so I is out and III stays.

Also, divide both equations, we get X^2b = 16 or x^b = 4, so II is true as well.
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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, 08:40
(X^a*X^b)/X^a/X^b=32/2
We get X^B=4

Since a,b,X are all positive we can work back and say A=2 and B=3 so both 2 and 3 are also correct. Therefore the 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]

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New post 15 Jul 2019, 08:46
Dividing statement 2/1

[x^(a+b)]/[x^(a-b)] = 32/2
x^2b = 2^4 ==> x^b = 4(statement 2 )

If we pre supposes a=3, b=2 and x=2 then the statement 1 and 3 also satisfies but here is a catch.
If we take the decimal values for a,b,x then statement 1 and 3 will not be satisfied.

Option 1 and 3 is maybe an option but NOT MUST.

only option 2 must be true.

Hence, B 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, 08:51
x^(a−b)=2
a-b=1

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

on solving a=3, b=2. I is correct.

II. x^b=4 = 2^2
b=2. Correct

III. b=(2/3)a
when a=3, b=2. Correct.

IMO 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, 08:53
IMO D.

Equation 1: x^(a−b)=2
Equation 2: x^(a+b)=32.

By multiplying these two equations we get, x^(a-b+a+b) = 2*32.
= x^(2a) = 64.

By dividing equation 2 by 1, we get, x^(a+b-a+b) = 32/2
= x^(2b) = 16.

x could take the values 2 or \sqrt{2},4.

If x=2, a= 3 and b=2 --> this satisfies x^(a-b) = 2 and x^(a+b) = 32.
If x= \sqrt{2}, a=6 and b=4. --> satisfies x^(a-b) = 2 and x^(a+b) = 32.
If x=4, b=1 but 'a' does not satisfy this equation.
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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] 15 Jul 2019, 08:53

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