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# If x is an integer, is 9^x + 9^{-x} = b ?

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Intern
Joined: 30 Oct 2016
Posts: 3
Re: If x is an integer, is 9^x + 9^{-x} = b ?  [#permalink]

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21 Feb 2018, 06:56
amanvermagmat wrote:
MintBlood wrote:
Bunuel wrote:
SOLUTION

If x is an integer, is $$9^x + 9^{-x} = b$$ ?

(1) $$3^x + 3^{-x} = \sqrt{b + 2}$$ --> square both sides --> $$9^x+2*3^x*\frac{1}{3^x}+9^{-x}=b+2$$ --> $$9^x + 9^{-x} = b$$. So answer to the question is YES. Sufficient.

(2) x > 0. No sufficient.

Could you please explain me how should I deal with the absolute value "b+2" after squaring both sides of the equation of option 2? I was confused by this part...
Really appreciate

Hi

Your question is not very clear. If you are talking about statement 1, there the right hand side is \sqrt{b + 2}[/m]
When squared, it will become (b + 2). Where does 'absolute value' come from?

If you are talking about statement 2, then it just tells us that x > 0. So [m]9^x + 9^{-x} can also be written as:
9^x + 1/9^x
But since we are not told anything about 'b' in statement 2, there is no way this statement is sufficient to answer this question.

Can you please explain what you meant by absolute value in this case?

Hi there, thanks for coming back.
Sorry that it should be statement 1 instead of 2.
I think more or less got myself the answer.
My question is previously (1) 3x+3−x=b+2‾‾‾‾‾√3x+3−x=b+2 --> square both sides --> 9x+2∗3x∗13x+9−x=b+29x+2∗3x∗13x+9−x=|b+2|, b+2 must be absolute |b+2|, which means the right side of the equition could be b+2 or -b-2....I suppose.
But since the first part " 9x+2∗3x∗13x+9−x" are all positive numbers, there is no way that |b+2| will be negative. I hope my logic here makes sense to you...
Anyway, thank you for ur answer
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Posts: 1348
Location: India
Re: If x is an integer, is 9^x + 9^{-x} = b ?  [#permalink]

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21 Feb 2018, 08:19
Could you please explain me how should I deal with the absolute value "b+2" after squaring both sides of the equation of option 2? I was confused by this part...
Really appreciate [/quote]

Hi

Your question is not very clear. If you are talking about statement 1, there the right hand side is \sqrt{b + 2}[/m]
When squared, it will become (b + 2). Where does 'absolute value' come from?

If you are talking about statement 2, then it just tells us that x > 0. So [m]9^x + 9^{-x} can also be written as:
9^x + 1/9^x
But since we are not told anything about 'b' in statement 2, there is no way this statement is sufficient to answer this question.

Can you please explain what you meant by absolute value in this case?[/quote]

Hi there, thanks for coming back.
Sorry that it should be statement 1 instead of 2.
I think more or less got myself the answer.
My question is previously (1) 3x+3−x=b+2‾‾‾‾‾√3x+3−x=b+2 --> square both sides --> 9x+2∗3x∗13x+9−x=b+29x+2∗3x∗13x+9−x=|b+2|, b+2 must be absolute |b+2|, which means the right side of the equition could be b+2 or -b-2....I suppose.
But since the first part " 9x+2∗3x∗13x+9−x" are all positive numbers, there is no way that |b+2| will be negative. I hope my logic here makes sense to you...
Anyway, thank you for ur answer [/quote]

Hi

In statement 1, when you square the RHS (right hand side), then the square of √(b+2) will be just (b+2) and NOT |b+2|.
Whenever we square √x, we get 'x', not |x|. Well its a different thing though that if we are taking √x (square root of x), then x must be non-negative, because square root of a negative quantity is not defined.

So in this question, statement 1, RHS is √(b+2) and its understood that (b+2) cannot be negative here, so when we take square of √(b+2) we will just write b+2.
Intern
Joined: 30 Oct 2016
Posts: 3
Re: If x is an integer, is 9^x + 9^{-x} = b ?  [#permalink]

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22 Feb 2018, 08:12
amanvermagmat wrote:
Could you please explain me how should I deal with the absolute value "b+2" after squaring both sides of the equation of option 2? I was confused by this part...
Really appreciate

Hi

Your question is not very clear. If you are talking about statement 1, there the right hand side is \sqrt{b + 2}[/m]
When squared, it will become (b + 2). Where does 'absolute value' come from?

If you are talking about statement 2, then it just tells us that x > 0. So [m]9^x + 9^{-x} can also be written as:
9^x + 1/9^x
But since we are not told anything about 'b' in statement 2, there is no way this statement is sufficient to answer this question.

Can you please explain what you meant by absolute value in this case?[/quote]

Hi there, thanks for coming back.
Sorry that it should be statement 1 instead of 2.
I think more or less got myself the answer.
My question is previously (1) 3x+3−x=b+2‾‾‾‾‾√3x+3−x=b+2 --> square both sides --> 9x+2∗3x∗13x+9−x=b+29x+2∗3x∗13x+9−x=|b+2|, b+2 must be absolute |b+2|, which means the right side of the equition could be b+2 or -b-2....I suppose.
But since the first part " 9x+2∗3x∗13x+9−x" are all positive numbers, there is no way that |b+2| will be negative. I hope my logic here makes sense to you...
Anyway, thank you for ur answer [/quote]

Hi

In statement 1, when you square the RHS (right hand side), then the square of √(b+2) will be just (b+2) and NOT |b+2|.
Whenever we square √x, we get 'x', not |x|. Well its a different thing though that if we are taking √x (square root of x), then x must be non-negative, because square root of a negative quantity is not defined.

So in this question, statement 1, RHS is √(b+2) and its understood that (b+2) cannot be negative here, so when we take square of √(b+2) we will just write b+2.
[/quote]

Ah! I see.... Thank you very much for clarifying! I can finally realise what has been going wrong in my logic...
Intern
Joined: 14 Feb 2016
Posts: 35
Re: If x is an integer, is 9^x + 9^{-x} = b ?  [#permalink]

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25 Apr 2018, 01:43
Review, synopsys:

Properties being tested: (square of a sum), (manipulation of equations with roots), and (operatons with exponents).

The steps I took to solve this problem are as follows:

Simplify the question stem to: 3^x + (⅓)^2x = b?

Let’s see what the GMAT is giving us:

3^x + 3^-x = Vb+2

Right away, we see that we’re dealing with a root on the right side. Let’s square it.

(3^x+3^-x)^2 = Vb+2

Remember to square the entire equation to get a square of a sum. Manipulate square of a sum accordingly.

3^2x + 3/3^x + 3/3^x + ⅓^2x

3 + 2 + (⅓)^2x = b+2

3^2x + (⅓)^2x = b

Hold on - this matches the question. So statement 1 is sufficient.

Statement 2 is insufficient.

Alternate solution:

We could have also solved this without rearrange the equation, and manipulation the first statement to match the question. But I find it helpful to always rearrange the question stem in a DS question with algebra.
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Posts: 92
Location: India
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Re: If x is an integer, is 9^x + 9^{-x} = b ?  [#permalink]

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23 Sep 2018, 03:10
Bunuel wrote:
SOLUTION

If x is an integer, is $$9^x + 9^{-x} = b$$ ?

(1) $$3^x + 3^{-x} = \sqrt{b + 2}$$ --> square both sides --> $$9^x+2*3^x*\frac{1}{3^x}+9^{-x}=b+2$$ --> $$9^x + 9^{-x} = b$$. So answer to the question is YES. Sufficient.

(2) x > 0. No sufficient.

(1) $$3^x + 3^{-x} = \sqrt{b + 2}$$ --> square both sides -->wht are we not taking both cases ie +-(3^x + 3^{-x})2=b+2
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Re: If x is an integer, is 9^x + 9^{-x} = b ? &nbs [#permalink] 23 Sep 2018, 03:10

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