Find the value of x.

Question

(1)  x = \sqrt{4096} 
(2)  x^{64} = 8^{128}

aggarwal1994 · Answer

Square root of any number is always positive. Hence, statement 1 gives x =64

In statement 2, x could be +64 or -64. Hence, it is not enough to arrive at a solution.

Ans: A

PGhind · Answer

This would be the case if we were solving for x^2= 64
but we aint doing that we are solving for x^64=8^128 which is equal to x=8^2
Please enlighten me if my logic is flawed.
Thank you

Chinmay1502 · Answer

According to Statement 1, x can be -64 or 64.
Same according to Statement 2.
Thus, answer should be E.
Right?

nousernamefine · Answer

1st is sq. root, so it cannot be -ve. Hence +64. Hope that helps

EshaFatim · Answer

I think the main confusion arrived with the first statement assumably yielding two results or not.

Statement 1:

We know that sqrt can give +ve and -ve. BUT... Look closely, you'll see the question has already given us the positive sign with the sqrt. So we don't need to worry about the negative sign.

If it was something like this -
 x^2=4096

Then it would yield two results +ve and -ve. But the process has been taken a step forward. What I mean is -

x^2=4096 
=>  x=+-\sqrt{4096}  ....... --> From this part they have eliminated the -ve root and shared with us the +ve root only

So we don't need to worry about the -ve root and the answer will simply be 64. Hope this helps.

Statement 2:

For statement 2 though, why can't we simply solve to x=64? +- shouldn't matter here, no? Both sides have positive even power.
 Bunuel , please help.

Bunuel · Answer

There should not be any confusion with this question.

From (1), since the square root cannot give a negative result, x can take only one value: 64.

From (2), because of the even root, we get two solutions. Taking the 64th root gives: |x| = 8^2 = 64, yielding x = 64 or x = -64 (similar to if we had x^2 = 4, which gives |x| = 2, so x = 2 or -2).

Answer: A.

anirchat · Answer

Hi Bunuel
The square root of a number can be negative. Every real number has two roots, one positive and one negative.
The correct answer should be E.

The only argument you can make is that 4096^1/2 has no sign before it, so we may assume that statement is considering the positive square root only which can make sense to have the answer as A.

Bunuel · Answer

You should brush up fundamentals.

Even roots cannot give negative result.

\sqrt{...}  is the square root sign, a function (called the principal square root function), which cannot give negative result. So,  this sign ( \sqrt{...} ) always means non-negative square root .

https://gmatclub.com/forum/download/file.php?id=39502 
 The graph of the function f(x) = √x

Notice that it's defined for non-negative numbers and is producing non-negative results.

TO SUMMARIZE:
  When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the non-negative root.   That is:

\sqrt{9} = 3 , NOT +3 or -3;
 \sqrt {16} = 2 , NOT +2 or -2;
Similarly  \sqrt{\frac{1}{16}} = \frac{1}{4} , NOT +1/4 or -1/4.

Notice that in contrast, the equation  x^2 = 9  has TWO solutions, +3 and -3. Because  x^2 = 9  means that  x =-\sqrt{9}=-3  or  x=\sqrt{9}=3 .

anirchat · Answer

Sure Bunuel. The notations differ from what we generally related use in India. But I guess what you are saying is the GMAT standard.

Bunuel · Answer

The GMAT doesn’t have its own specific math rules; what’s stated above is generally true.

anirchat · Answer

Understood. Thanks for your prompt response.

Find the value of x.

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