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If x^(ab) = 2 and x^(a+b) = 32, and if x, a, and b are greater than
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15 Jul 2019, 08:00
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If \(x^{(ab)} = 2\) 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^(ab) = 2 and x^(a+b) = 32, and if x, a, and b are greater than
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Updated on: 16 Jul 2019, 06:28
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 \(x^(ab) =2\) (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^(ab) = 2 and x^(a+b) = 32, and if x, a, and b are greater than
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15 Jul 2019, 08:28
\(x^{(a−b)}=2\) 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^(ab) = 2 and x^(a+b) = 32, and if x, a, and b are greater than
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15 Jul 2019, 08:12
1. x^(ab)=2^1 > ab=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^(ab) = 2 and x^(a+b) = 32, and if x, a, and b are greater than
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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^(ab) = 2 and x^(a+b) = 32, and if x, a, and b are greater than
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15 Jul 2019, 08:15
solving, 5(ab)=(a+b) gives a= 3b/2 using this x^b=4



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If x^(ab) = 2 and x^(a+b) = 32, and if x, a, and b are greater than
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Updated on: 15 Jul 2019, 09:52
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 \((ab) =1\). As a valid option X could be \(\sqrt{2}\) and \((ab) =2\), 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 \(=> [(ab) = (a+b)/5] => [(a+b) = 5(ab)] => [6b=4a] =>[b = (\frac{2}{3})*a]\), which is 3rd statement.
Now, lets assume \(ab = y\), then \(a+b = 5y\) (see above) \(=> [ab  (a+b) = y5y] => [2b= 4y] => [b=2y]\). 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^(ab) = 2 and x^(a+b) = 32, and if x, a, and b are greater than
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15 Jul 2019, 08:22
Answer is E. I, II, and III



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Re: If x^(ab) = 2 and x^(a+b) = 32, and if x, a, and b are greater than
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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^(ab) = 2 and x^(a+b) = 32, and if x, a, and b are greater than
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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^ab=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^(ab) = 2 and x^(a+b) = 32, and if x, a, and b are greater than
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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 = 5a5b => 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^(ab) = 2 and x^(a+b) = 32, and if x, a, and b are greater than
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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^(ab) = 2 and x^(a+b) = 32, and if x, a, and b are greater than
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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^(ab) = 2 and X^(a+b) = 32 we can write x^(a+b) = x^(5*(ab)) simplifying the two sides we get a+b = 5 (ab) 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^(ab) = 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^(ab) = 2 and x^(a+b) = 32, and if x, a, and b are greater than
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Updated on: 15 Jul 2019, 23:29
x^(a−b)=2 x^(a+b)=32=2^5 = (x^(ab))^5 x^(a+b) = (x^5(ab)) a+b=5(ab) 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
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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^(ab) = 2 and x^(a+b) = 32, and if x, a, and b are greater than
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15 Jul 2019, 08:38
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 root2^4III. 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^(ab) = 2 and x^(a+b) = 32, and if x, a, and b are greater than
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15 Jul 2019, 08:39
D
From the given equations: x^(a−b)=2, and x^(a+b)=32 = 2^5. Comparing both equations, we get 5*(ab) = 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^(ab) = 2 and x^(a+b) = 32, and if x, a, and b are greater than
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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^(ab) = 2 and x^(a+b) = 32, and if x, a, and b are greater than
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15 Jul 2019, 08:46
Dividing statement 2/1
[x^(a+b)]/[x^(ab)] = 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^(ab) = 2 and x^(a+b) = 32, and if x, a, and b are greater than
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15 Jul 2019, 08:51
x^(a−b)=2 ab=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^(ab) = 2 and x^(a+b) = 32, and if x, a, and b are greater than
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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^(ab+a+b) = 2*32. = x^(2a) = 64.
By dividing equation 2 by 1, we get, x^(a+ba+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^(ab) = 2 and x^(a+b) = 32. If x= \sqrt{2}, a=6 and b=4. > satisfies x^(ab) = 2 and x^(a+b) = 32. If x=4, b=1 but 'a' does not satisfy this equation.




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