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

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The Official Guide for GMAT® Review, 13th Edition - Quantitative Questions Project

If x is an integer, is \(9^x + 9^{-x} = b\) ?

(1) \(3^x + 3^{-x} = \sqrt{b + 2}\)
(2) x > 0

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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.

Answer: A.
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Re: If x is an integer, is 9^x + 9^{-x} = b ? [#permalink]

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1) \(3^x + 3^-^x = \sqrt{b + 2}\)

square both sides

\(9^x+2*3^x*\frac{1}{3^x}+9^{-x} = b+2\)

Therefore, \(9^x + 9^{-x} = b\).

Sufficient.

2) Tells us nothing about b, but rather that x is a positive number. Insufficient.

[Reveal] Spoiler:
Knowing this, the solution is A - statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.

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Re: If x is an integer, is 9^x + 9^{-x} = b ? [#permalink]

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This was tough. I spent at least 3 minutes (during the exam is not possible) :(

We can rephrase the stimulus but I see a lot of options to do. Is not straight (at least for me). Ok better to look at the statements

2) this say nothing about \(=\)between the left part of equation and the right part INSUFF

1) square boot sides so we have \((3^x + 3^-x)^2\)\(=\)\(\sqrt{b + 2}^2\)

Now we 'd have \(9^x\) that is, is the same of \(3^2x\) -------> \(9^x + 9^-x + 2 ( 3^x + 3^-x)\)\(=\) \(b + 2\) ---------> \(9^x + 2 + 9^-x = b + 2\)

In the end \(9^x + 9^-x = b\) SUFF

A should be the answer

Note: \(2 ( 3^x + 3^-x)\) is zero because we have \(3^ x- x\) . a number power zero is 1 ---> \(2*1 = 2\) for me is more than 600 level
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Re: If x is an integer, is 9^x + 9^{-x} = b ? [#permalink]

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1) Square both the side & get the same equation as in stem --->Sufficient
2) Insufficient
Answer A
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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.

Answer: A.

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Re: If x is an integer, is 9^x + 9^{-x} = b ? [#permalink]

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New post 31 Oct 2012, 23:40
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.

Answer: A.

Kudos points given to everyone with correct solution. Let me know if I missed someone.


Question please: Can we do anything with the 9^x + 9^{-x} = b or simplify any more than what is given? I tried to do something more but could not find anything proper...

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Re: If x is an integer, is 9^x + 9^{-x} = b ? [#permalink]

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New post 01 Nov 2012, 07:25
ikokurin 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.

Answer: A.

Kudos points given to everyone with correct solution. Let me know if I missed someone.


Question please: Can we do anything with the 9^x + 9^{-x} = b or simplify any more than what is given? I tried to do something more but could not find anything proper...


I'd say \(9^x + 9^{-x}\) the simplest way of writing this expression and as you can see from the solution we don't even need to manipulate with it further to answer the question.
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Re: If x is an integer, is 9^x + 9^{-x} = b ? [#permalink]

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bb, I tried navigating below link, but it says "You are not authorised to read this forum". Can you check please?

the-official-guide-for-gmat-review-13th-edition-quant-134495.html
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why is 3^x * 3^x = 9x? shouldnt it be 9^2x?
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Re: If x is an integer, is 9^x + 9^{-x} = b ? [#permalink]

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New post 17 Nov 2014, 15:24
As was noted earlier in this post, this question uses the formula (x+y)^2=x^2+2xy+y^2. In this question xy=1 because x^0=1. This allows for the 2's to cancel out.

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Re: If x is an integer, is 9^x + 9^{-x} = b ? [#permalink]

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New post 05 May 2015, 15:02
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.

Answer: A.


When you square both sides, I don't understand where the +2 comes from? Can you please explain?

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Re: If x is an integer, is 9^x + 9^{-x} = b ? [#permalink]

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New post 05 May 2015, 15:53
dutchmen991 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.

Answer: A.


When you square both sides, I don't understand where the +2 comes from? Can you please explain?


Hello dutchmen991
this is formula that used to square expression:
\((a+b)^2 = (a+b)(a+b) = a^2+2ab+b^2\)

In our case we have
\(3^x + 3^{-x}=3^x + \frac{1}{3^{x}}\)
When we square this expression we will have:
\((3^x + \frac{1}{3^{x}})*(3^x + \frac{1}{3^{x}})=9^x+3x∗\frac{1}{3^x}+\frac{1}{3^x}*3^x+\frac{1}{9^x}=9^x+2*3^x*\frac{1}{3^x}+9^{-x}\)
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If x is an integer, is 9^x + 9^{-x} = b ? [#permalink]

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New post 05 May 2015, 16:04
Harley1980 wrote:
dutchmen991 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.

Answer: A.


When you square both sides, I don't understand where the +2 comes from? Can you please explain?


Hello dutchmen991
this is formula that used to square expression:
\((a+b)^2 = (a+b)(a+b) = a^2+2ab+b^2\)

In our case we have
\(3^x + 3^{-x}=3^x + \frac{1}{3^{x}}\)
When we square this expression we will have:
\((3^x + \frac{1}{3^{x}})*(3^x + \frac{1}{3^{x}})=9^x+3x∗\frac{1}{3^x}+\frac{1}{3^x}*3^x+\frac{1}{9^x}=9^x+2*3^x*\frac{1}{3^x}+9^{-x}\)



Thanks for the reply. I didn't realize this was difference of squares because I didn't know how to deal with the negative exponent. Is this the same concept being tested in the following two problems?

is-5-k-less-than-144719.html
if-2-x-2-x-2-3-2-13-what-is-the-value-of-x-130109.html

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Re: If x is an integer, is 9^x + 9^{-x} = b ? [#permalink]

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New post 05 May 2015, 16:21
dutchmen991 wrote:

Thanks for the reply. I didn't realize this was difference of squares because I didn't know how to deal with the negative exponent. Is this the same concept being tested in the following two problems?

is-5-k-less-than-144719.html
if-2-x-2-x-2-3-2-13-what-is-the-value-of-x-130109.html


Yeah, all this tasks tests understanding of work with exponents/powers.
You can search them by tag:
search.php?search_id=tag&tag_id=60
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Re: If x is an integer, is 9^x + 9^{-x} = b ? [#permalink]

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New post 22 May 2016, 15:24
Thought process here:

question is asking : is 9^x + 9^-x = b?

so we need to figure out how to make this happen from the data

1) 3^x + 3^-x = √b+2)
square both sides
(3^x + 3^-x)(3^x + 3^-x) = b+2
9^x + 9^-x + 2 = b + 2
subtract 2 = 9^x + 9^-x = b. SUFFICIENT

2)x > 0
This could give you multiple answers, so cannot work

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Re: If x is an integer, is 9^x + 9^{-x} = b ? [#permalink]

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New post 22 May 2016, 15:24
Thought process here:

question is asking : is 9^x + 9^-x = b?

so we need to figure out how to make this happen from the data

1) 3^x + 3^-x = √b+2)
square both sides
(3^x + 3^-x)(3^x + 3^-x) = b+2
9^x + 9^-x + 2 = b + 2
subtract 2 = 9^x + 9^-x = b. SUFFICIENT

2)x > 0
This could give you multiple answers, so cannot work

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Re: If x is an integer, is 9^x + 9^{-x} = b ?   [#permalink] 22 May 2016, 15:24

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