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

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

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01 Oct 2012, 05:20
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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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Re: If x is an integer, is 9^x + 9^{-x} = b ?  [#permalink]

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01 Oct 2012, 05:20
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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.

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

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01 Oct 2012, 05:40
1
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.

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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01 Oct 2012, 05:54
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1
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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01 Oct 2012, 20:18
1
1) Square both the side & get the same equation as in stem --->Sufficient
2) Insufficient
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Re: If x is an integer, is 9^x + 9^{-x} = b ?  [#permalink]

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04 Oct 2012, 14:44
1
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.

Kudos points given to everyone with correct solution. Let me know if I missed someone.
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Re: If x is an integer, is 9^x + 9^{-x} = b ?  [#permalink]

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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.

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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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.

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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13 Nov 2012, 18:05
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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14 Nov 2012, 02:51
1
JJ2014 wrote:
why is 3^x * 3^x = 9x? shouldnt it be 9^2x?

$$3^x * 3^x=3^{x+x}=3^{2x}=9^x$$.

For more check Number Theory chapter of our Math Book: math-number-theory-88376.html

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

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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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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.

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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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.

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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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.

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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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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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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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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16 Jul 2017, 14:46
1
If x is an integer, is $$9^x + 9^{-x} = b$$ ?

$$3^{2x} + 3^{-2x} = b$$

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

$$3^x + 3^{-x} = \sqrt{b + 2}$$

Squaring on both sides:

$$(3^x + 3^{-x})^2 = b + 2$$

$$3^{2x} + 2 * 3^{x} * 3^{-x} + 3^{-2x} = b + 2$$

$$3^{2x} + 2 + 3^{-2x} = b + 2$$

$$3^{2x} + 3^{-2x} = b$$

This matches to the above-simplified equation of the question.

Hence, (1) ===== is SUFFICIENT

(2) x > 0

Clearly Not Sufficient

Hence, (2) ===== is NOT SUFFICIENT

Hence, Answer is A

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

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19 Feb 2018, 06:30
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
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Re: If x is an integer, is 9^x + 9^{-x} = b ?  [#permalink]

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19 Feb 2018, 09:11
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?
Re: If x is an integer, is 9^x + 9^{-x} = b ? &nbs [#permalink] 19 Feb 2018, 09:11

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