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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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Question Stats: 73% (01:48) correct 27% (01:58) wrong based on 1719 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

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Question: 53
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Difficulty: 600

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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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5
11
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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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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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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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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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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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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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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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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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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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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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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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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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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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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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

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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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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 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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