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

\(x^{(a−b)}=2\) 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)

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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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 \(x=\sqrt[30]{2}\).
If \(b=60\), then \(x^b=4\) or \(x^{60}=4\) or \(x=\sqrt[30]{2}\). We can see that \(a=90\) when \(x=\sqrt[30]{2}\)

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

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

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

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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