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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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$$x^{(a−b)}=2$$ 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

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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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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.

Posted from my mobile device

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

Originally posted by JonShukhrat on 15 Jul 2019, 09:03.
Last edited by JonShukhrat on 15 Jul 2019, 21:37, 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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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)

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
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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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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

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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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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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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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