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If x is an integer, is ....

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by factor26 » Mon Dec 19, 2011 6:32 pm

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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

I have no idea where to even start here... This looks like one of those questions I would make an educated guess on and then just quickly move on....can someone please give me a background as to what this question is asking for and how to solve?

Og answer is A --- Tks all!
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Source: — Data Sufficiency |
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by GmatMathPro » Mon Dec 19, 2011 7:07 pm
factor26 wrote:If x is an integer is 9^ x + 9^-x =b?

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

2- x>0

I have no idea where to even start here... This looks like one of those questions I would make an educated guess on and then just quickly move on....can someone please give me a background as to what this question is asking for and how to solve?

Og answer is A --- Tks all!
Did you try squaring both sides of statement 1? It works out more nicely than you might think.
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by Anurag@Gurome » Mon Dec 19, 2011 7:14 pm
factor26 wrote:If x is an integer is 9^ x + 9^-x =b?

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

2- x>0

I have no idea where to even start here... This looks like one of those questions I would make an educated guess on and then just quickly move on....can someone please give me a background as to what this question is asking for and how to solve?

Og answer is A --- Tks all!

(1) Square both sides of 3^x + 3^(-x) = √(b + 2)
3^(2x) + 3^(-2x) + 2[3^x * 3^(-x)] = b + 2
9^x + 9^(-x) + 2[3^(x - x)] = b + 2
9^x + 9^(-x) + 2[3^0] = b + 2
9^x + 9^(-x) + 2 = b + 2 (since 3^0 = 1)
So, 9^x + 9^(-x) = b; SUFFICIENT.

(2) x > 0 does not give us any information about b, so statement 2 is NOT sufficient.

The correct answer is A.
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by GMATGuruNY » Mon Dec 19, 2011 7:41 pm
factor26 wrote:If x is an integer is 9^ x + 9^-x =b?

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

2- x>0

I have no idea where to even start here... This looks like one of those questions I would make an educated guess on and then just quickly move on....can someone please give me a background as to what this question is asking for and how to solve?

Og answer is A --- Tks all!
Many test-takers will find the algebra here daunting.
An alternate approach is to plug in an easy number for x and solve for b.

Statement 1: 3^x + 3^(-x) = √(b+2).
Let x=1.
Then:
3¹ + 3ˉ¹ = √(b+2)
3 + 1/3 = √(b+2)
10/3 = √(b+2).

Squaring both sides:
100/9 = b+2
b = 100/9 - 2 = 100/9 - 18/9 = 82/9.

Does 9^ x + 9^-x = b?
Plugging x=1 and b=82/9 into the question stem, we get:
9¹ + 9ˉ¹ = 82/9
9 + 1/9 = 82/9
81/9 + 1/9 = 82/9
82/9 = 82/9.
YES.
Not a definitive proof, but we can be reasonably certain that statement 1 tells us that 9^ x + 9^-x = b.
SUFFICIENT.

Statement 2: x>0.
No information about b.
INSUFFICIENT.

The correct answer is A.
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by factor26 » Tue Dec 20, 2011 2:34 pm
Thanks all!!

@gmatguruny -- is it ok to use simple numbers, #1 for the value of X?

I thought one should steer clear of using the number 1 when plugging solving such questions like this.
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by GMATGuruNY » Wed Dec 21, 2011 6:34 am
factor26 wrote:Thanks all!!

@gmatguruny -- is it ok to use simple numbers, #1 for the value of X?

I thought one should steer clear of using the number 1 when plugging solving such questions like this.
When a PS question has variables in the answer choices, we should avoid plugging in 1.
The reason is that plugging in 1 increases the odds that more than one answer choice will work.

To illustrate:
Let's say that, in response to a given PS problem, we plug in x=1 and get a target of 5.
Consider the following answer choices:

5/x
5/x²
5x
5x²
5^x

When x=1, every answer choice is equal to 5.
The result: we have to plug in a different value for x and repeat the whole process.

DS problems don't have 5 answer choices, each containing a variable.
Instead, we're evaluating the 2 statements, so our concern here is different.
We need to be mindful that the result yielded by x=1 might not be the same as the result that would be yielded by another value.
In the DS problem above, plugging in x=1 makes the math SO much more manageable that the benefits seem to outweigh the risk.
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by dario.brignone » Sat Mar 03, 2012 6:49 am
Without calculating anything, could we simply say that A is sufficient because we have two variables and two equations?
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by kitg » Sat Mar 03, 2012 7:50 am
i guess we can directly say this!
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by r18ory » Thu Jun 06, 2013 2:05 am
Hi this is my first post, which shows how much difficulty I am having with this problem. My difficulty comes from the rule of exponents that (x^r)^2 = x^2r. If this is the case then surely:

(3^x + 3^-x)^2 = 3^2x + 3^-2x

However in all solutions above and in the OG the result of (3^x + 3^-x)^2 is:

3^2x + 3^-2x + 2(3^x + 3^-x)

Where does the additional 2(3^x + 3^-x) come from? I have spent the last 2 hours staring at this question and appear to understand all exponent rules on https://www.purplemath.com/modules/simpexpo2.htm, but I just can't work out where the 2(3^x + 3^-x) comes from if the (x^r)^2 = x^2r rule is to be believed!
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by Atekihcan » Thu Jun 06, 2013 3:34 am
r18ory wrote:then surely:

(3^x + 3^-x)^2 = 3^2x + 3^-2x

However in all solutions above and in the OG the result of (3^x + 3^-x)^2 is:

3^2x + 3^-2x + 2(3^x + 3^-x)
Both are wrong.
(a + b)² = a² + b² + 2ab

So, (3^x + 3^-x)^2
= (3^x)^2 + (3^-x)^2 + 2*(3^x)*(3^-x)
= 3^2x + 3^-2x + 2*(3^(x - x))
= 3^2x + 3^-2x + 2*(3^0)
= 3^2x + 3^-2x + 2
Last edited by Atekihcan on Thu Jun 06, 2013 3:35 am, edited 1 time in total.
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by GMATGuruNY » Thu Jun 06, 2013 3:35 am
r18ory wrote:Hi this is my first post, which shows how much difficulty I am having with this problem. My difficulty comes from the rule of exponents that (x^r)^2 = x^2r. If this is the case then surely:

(3^x + 3^-x)^2 = 3^2x + 3^-2x

However in all solutions above and in the OG the result of (3^x + 3^-x)^2 is:

3^2x + 3^-2x + 2(3^x + 3^-x)

Where does the additional 2(3^x + 3^-x) come from? I have spent the last 2 hours staring at this question and appear to understand all exponent rules on https://www.purplemath.com/modules/simpexpo2.htm, but I just can't work out where the 2(3^x + 3^-x) comes from if the (x^r)^2 = x^2r rule is to be believed!
Your presumption that (x+y)² = x² + y² is incorrect.
Consider the following case:
(2+3)² = 2² + 3²
25 = 13.
Doesn't work.

Memorize the following quadratic indentities:
(a+b)² = a² + 2ab + b²
(a-b)² = a² - 2ab + b²
(a+b)(a-b) = a² - b².

The identity in red is relevant here:
(a+b)² = a² + 2ab + b²
(3^x + 3 ^-x)² = (3^x)² + 2(3^x)(3^-x) + (3^-x)².

Simplifying further, we get:
(3^x)² + 2(3^x)(3^-x) + (3^-x)²

= 3^(2x) + 2(3^0) + 3^(2*-x)

= 9^x + 2 + 9^(-x).
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by Scott@TargetTestPrep » Mon Sep 17, 2018 5:39 pm
factor26 wrote:If x is an integer is 9^ x + 9^-x =b?

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

2- x>0
We need to determine whether 9^x + 9^-x = b.

Statement One Alone:

3^x + 3^-x = √(b+2)

We first square both sides of the equation in statement one, obtaining:

(3^x + 3^-x)^2 = [√(b+2)]^2

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

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

3^(2x) + 3^0 + 3^0 + 3^(-2x) = b + 2

9^x + 1 + 1 + 9^-x = b + 2

9^x + 9^-x = b

Statement one is sufficient to answer to the question.

Statement Two Alone:

x> 0

Knowing only that x is greater than zero is not enough information to answer the question.

Answer: A

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