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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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New post 01 Oct 2012, 05:20
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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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New post 01 Oct 2012, 05:20
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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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New post 01 Oct 2012, 05:40
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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.

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

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

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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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New post 19 Aug 2014, 22:35
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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.

Answer (A)
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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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New post 05 Dec 2015, 12:56
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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Re: If x is an integer, is 9^x + 9^(-x) = b ? (1) 3^x + 3^(-x) = (b + 2)  [#permalink]

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New post 16 Jul 2017, 14:46
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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 ? (1) 3^x + 3^(-x) = (b + 2)  [#permalink]

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New post 06 Feb 2019, 04:24
when we square 3^x+3^-x the result should not be 9^x^2+9^-x^2? thank you in advance
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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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New post Updated on: 08 Apr 2019, 06:46
Bunuel VeritasKarishma or other experts, is this correct?

Stem:
9^x+9^ -x = b?
(3^2)^x+(3^2)^-x = b?
3^2(1^x+1^-x)
3^2(2) =
18 = b?

1) 3^x+3^-x = √(b+2)
3(1^x+1^-x) = (3^x+3^-x)^2 = b+2
3(2) = 3^2x + 3^0 + 3^0 + 3^-2x = b+2
36 = b+2 3^2x + 3^-2x + 2 = b + 2
34 = b 3^2x+3^-2x = b
Definite yes, Sufficient

2) Irrelevant because base 1 will always be 1 if x is an int.

Edit: Thanks Karishma, I must have been half asleep when I did this one...
Takeaway: Don't confuse multiplication and addition of exponents when factoring out!

Originally posted by energetics on 05 Apr 2019, 13:48.
Last edited by energetics on 08 Apr 2019, 06:46, edited 3 times in total.
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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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New post 05 Apr 2019, 22:52
energetics wrote:
Bunuel VeritasKarishma or other experts, is this correct?

Stem:
9^x+9^ -x = b?
(3^2)^x+(3^2)^-x =
3^2(1^x+1^-x) =
3^2(2) =
18 = b?

1) 3^x+3^-x = √(b+2)
3(1^x+1^-x) =
3(2) =
36 = b+2
34 = b
Definite no, Sufficient

2) Irrelevant because base 1 will always be 1 if x is an int.


No energetics.

Note that we cannot add the exponents when the terms are added. We do that only when the terms are multiples.

3^a * 3^b = 3^(a + b) ---- Correct

3^a + 3^b is not 3^(a + b).
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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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New post 05 Apr 2019, 23:42
Bunuel wrote:
If x is an integer, is \(9^x + 9^{-x} = b\) ?


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

(2) x > 0



Question : is \(9^x + 9^{-x} = b\)


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

i.e. Squaring both sides \(3^{2x} + 3^{-2x} +2*(3^x*3^{-x} = √(b+2)\)

i.e. \(3^{2x} + 3^{-2x} +2 = (b+2)\)

i.e. \(9^x + 9^{-x} = (b)\)

i.e. answer to the question is DEFINITELY YES hence

SUFFICIENT

Statement 2: x > 0

No association of x with b is given to check the question statement's authenticity to answer the question hence

NOT SUFFICIENT

Answer: Option A
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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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New post 04 Jun 2019, 06:01
I'm confused ...
3^x * 3^x should be equal to 3^2x not 9^x
if answer is 9^x it must to be (3^2)^x....
it's totally different
can someone explain
thank in advance.....
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Re: If x is an integer, is 9^x + 9^(-x) = b ? (1) 3^x + 3^(-x) = (b + 2)   [#permalink] 04 Jun 2019, 06:01
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