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

Question

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

Kinshook · Accepted Answer

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

berdibekov · Answer

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:
   =>   , since X>0  =>   =>   =>   =>  , which is 3rd statement.

Now, lets assume  a-b = y , then  a+b = 5y  (see above)  =>   =>   =>   . 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

firas92 · Answer

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)

prashanths · Answer

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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JonShukhrat · Answer

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 {2} . 
If  b=60 , then  x^b=4  or  x^{60}=4  or  x=\sqrt {2} . We can see that  a=90  when  x=\sqrt {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

nick1816 · Answer

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

Abhi077 · Answer

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

sj24 · Answer

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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