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# If x is an integer, is 9^x + 9^(-x) = b ? (1) 3^x + 3^(-x) = (b + 2)

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If x is an integer, is 9^x + 9^(-x) = b ? (1) 3^x + 3^(-x) = (b + 2)  [#permalink]

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01 Oct 2012, 04:20
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Difficulty:

35% (medium)

Question Stats:

74% (01:19) correct 26% (01:29) wrong based on 1701 sessions

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

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

(2) x > 0

Practice Questions
Question: 53
Page: 279
Difficulty: 600

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

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Re: If x is an integer, is 9^x + 9^(-x) = b ? (1) 3^x + 3^(-x) = (b + 2)  [#permalink]

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01 Oct 2012, 04:20
3
10
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 ? (1) 3^x + 3^(-x) = (b + 2)  [#permalink]

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01 Oct 2012, 04:40
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 ? (1) 3^x + 3^(-x) = (b + 2)  [#permalink]

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01 Oct 2012, 04:54
2
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

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 ? (1) 3^x + 3^(-x) = (b + 2)  [#permalink]

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31 Oct 2012, 22:40
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 ? (1) 3^x + 3^(-x) = (b + 2)  [#permalink]

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01 Nov 2012, 06:25
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 ? (1) 3^x + 3^(-x) = (b + 2)  [#permalink]

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13 Nov 2012, 17:05
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 ? (1) 3^x + 3^(-x) = (b + 2)  [#permalink]

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14 Nov 2012, 01:51
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 ? (1) 3^x + 3^(-x) = (b + 2)  [#permalink]

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19 Aug 2014, 21:35
1
2
yb wrote:
If x is an integer, is 9^x + 9^(-x) = b ?

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

2) x>0

Note that 9^x is the square of 3^x so we know that we should try to square stmnt 1.

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

We answer with 'Yes' and hence this statement alone is sufficient.

2) x > 0
Obviously not sufficient.

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Re: If x is an integer, is 9^x + 9^(-x) = b ? (1) 3^x + 3^(-x) = (b + 2)  [#permalink]

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05 Dec 2015, 11:56
Bunuel , what is x=0 ?
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Re: If x is an integer, is 9^x + 9^(-x) = b ? (1) 3^x + 3^(-x) = (b + 2)  [#permalink]

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06 Dec 2015, 09:31
saroshgilani wrote:
Bunuel , what is x=0 ?

If x = 0, then the questions asks whether b = 2.

(1) gives $$2 = \sqrt{b+2}$$ --> b = 2. Sufficient.

Hope it's clear.
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Re: If x is an integer, is 9^x + 9^(-x) = b ? (1) 3^x + 3^(-x) = (b + 2)  [#permalink]

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16 Jul 2017, 13:46
2
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

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Re: If x is an integer, is 9^x + 9^(-x) = b ? (1) 3^x + 3^(-x) = (b + 2)  [#permalink]

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26 Nov 2018, 23:56
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