If x is a positive integer, is x greater than 4?

Question

If  x  is a positive integer, is  x > 4 ?

(1)  x^2 + 32 = 12x

(2)  x^x = \sqrt{x^{16}}

IanStewart · Accepted Answer

Solving the quadratic in Statement 1, x can be 4 or 8, so we don't know if x > 4,

From Statement 2, x^x = x^8. We have equal bases on each side, so if the bases are greater than 1, the exponents need to be equal, and x = 8. But it's also possible that x = 1, because 1^1 = 1^8, and again we can't tell if x > 4.

Using both Statements, x = 8, so the answer is C.

PyjamaScientist · Answer

From first statement you get two values of x (4 and 8). This is insufficient to answer ifx>4.

From second statement, x=8. With this you can sufficiently answer that x is greater than 4 since it is equal to 8.

Hence option B is the correct choice.

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PyjamaScientist · Answer

That's the mistake I often make. Not factoring in all the possibilities. Missed the part where X could be equal to 1.

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BrentGMATPrepNow · Answer

Given :  x  is a positive integer

Target question :    Is  x > 4 ?

Statement 1 :  x^2 + 32 = 12x   
Since we are given a quadratic equation let's first set it equal to zero:  x^2 - 12x + 32 = 0 
Factor to get:  (x - 4)(x - 8) = 0 , which means EITHER  x = 4  OR  x =8 . 
 Case a : If  x = 4 , the answer to the target question is  NO, x is not greater than 4 
 Case b : If  x = 8 , the answer to the target question is  YES, x is greater than 4 
Since we can’t answer the  target question  with certainty, statement 1 is NOT SUFFICIENT

Statement 2 :  x^x = \sqrt{x^{16}}  
Rewrite the equation as follows:  x^x = (x^{16})^{\frac{1}{2}}   - Taking the square root of a value is equivalent to raising that value to the power of  \frac{1}{2}  
Simplify the right side to get:  x^x = x^8

Note: I created this question to remind students that, when it comes to equations with variables in the exponents, there are three important provisos we must consider before we can conclude that two exponents are equal. 
That is, if  x^a = x^b , then we can conclude that  a = b    AS LONG AS    x 
eq 0 ,  x 
eq 1 , and  x 
eq -1 . 
For example, if we know that  0^x = 0^3 , we can't then conclude that  x = 3 , since there are infinitely many values of  x  that satisfy the equation.

So, if  x  does NOT equal -1, 0 or 1, then we can conclude that  x = 8

Now let's test the forbidden numbers (i.e., -1, 0 and 1).
Since we're told  x  is positive, we need only test whether  x = 1  satisfies the given equation. 
Plug  x = 1  into the equation  x^x = x^8  to get:  1^1 = 1^8 ....IT WORKS!!
So,  x = 1  is another solution to the equation.

Case a : If  x = 8 , the answer to the target question is  YES, x is greater than 4 
 Case b : If  x = 1 , the answer to the target question is  NO, x is not greater than 4 
Since we can’t answer the  target question  with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined    
Statement 1 tells us that either  x = 4  or  x =8 
Statement 2 tells us that either  x = 1  or  x =8 
Since  x =8  is the only x-value the two statements have in common, we can be certain that  x =8 , which means the answer to the target question is  YES, x is greater than 4 
Since we can answer the  target question  with certainty, the combined statements are SUFFICIENT

Answer: C

parth04red · Answer

Given: x>0

To find: x>4?

St (1): x^2 + 32 = 12x
 x^2 - 12x + 32 = 0
 (x-4)(x-8) = 0
 thus, x = 4 (not greater than 4) {ans to our original question:  NO }
 x = 8 { YES }
 
  Not Sufficient

St (2): x^x = \sqrt{x^16}
x^2x = x^16
this essentially equates to x = 8, which would be sufficient to answer the original question.
but if x = 1,
x^2x = x^16 = 1
thus, if x = 1 { NO }
 x = 8 { YES }

Not Sufficient

St 1 and 2 taken together
only possible value of x is 8 ( 8>4 )
  Sufficient

Answer:  option C

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If x is a positive integer, is x greater than 4?

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If x is a positive integer, is x greater than 4?

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