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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

2

This post received KUDOS

1

This post was BOOKMARKED

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 _________________

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.

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...

(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. _________________

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

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07 Jan 2014, 08:43

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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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.

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.

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

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.

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?

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?

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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