If x^2 = 2^x, what is the value of x ?

Question

If  x^2 = 2^x , what is the value of x ?

(1)  2x = (\frac{x}{2})^3

(2)  x = 2^{x-2}

chetan2u · Accepted Answer

Hi..

If  x^2 = 2^x , what is the value of x ?

(1)  2x = (\frac{x}{2})^3 
 x*(\frac{x^2}{8}-2)=0 
 we get x as 0, 4 or -4..
 when we substitute value in  x^2 = 2^x , ONLY x = 4 is possible..
sufficient

(2)  x = 2^{x-2} 
or  x=2^{x-2} ...
 2^x has to be multiple of 2 and so also x should be in power of 2 that is 2,4,8,16...AND x will fit in only at x=4 as  2^{4-2} 
ONLY when x is 4 both sides are 4. .
suff

D

carcass   i believe statement II should be  x=2^{x-2} .. here possible value is 4 SAME as statement I.. pl relook into Q

Bunuel · Answer

First of all, notice that (2) is  x = 2^{x-2}  (2 in power of x-2) not  x = 2^x-2 .

x = 2^{x-2} ;

x = \frac{2^x}{4} ;

Since  x^2 = 2^x , then  x = \frac{x^2}{4} ;

x(\frac{x}{4}-1)=0 ;

x = 0 or x = 4.

Only x = 4 satisfies  x^2 = 2^x , this x = 4. Sufficient.

Hope it's clear.

manishtank1988 · Answer

My 2 cents: my 2 cents.png

cocojatti92 · Answer

hey,
Still confused on statement 2
x=2^x-2
plugged that into x^2=2^x
2^(2x)-2^(2)=(2^x-2^2)
2x-2=2x-2
where I'm going wrong

dave13 · Answer

Bunuel how did you get highlighted part ? Re: st 2 i plugged in x = 2^{x-2} in x^2 = 2^x , and got 2^{(x-2)}^{2} = 2^{2^{x-2}} 2^{2x-4} = 2^{2^{x-2}} equate bases ---> 2x - 4 = 2^{x-2} 2x - 2^2 = 2^{x-2} 2x = 2^{x-2} + 2^2 so i got stuck here how to correctly factor out 2

Any idea ?

Bunuel · Answer

x = 2^{x-2} ;

x = 2^{x}*2^{-2} ;

x = 2^{x}*\frac{1}{2^{2}} ;

x = \frac{2^x}{4} .

WheatyPie · Answer

Nowhere is it stated that x is an integer. There is an irrational solution to statement 2 where x is approximately equal to 0.309907.

As posed, statement 2 is not sufficient. The question should be rewritten to say that x is an integer.

IanStewart · Answer

No, that is not the case - you're not using the information in the question stem, that  x^2 = 2^x

Here when analyzing Statement 2, if we notice that x must be nonzero, we can divide the equation in the stem by that in Statement 2:

\begin{align}
x^2 &= 2^x \
x &= 2^{x-2}
\end{align}

to get

x = 2^2 = 4

so Statement 2 is sufficient.

WheatyPie · Answer

You're right, I was forgetting about the stem. However, the stem also admits non-integer solutions. It would be difficult (and also out of scope) to check whether the irrational solutions to the stem coincide with those of statement 2.

It's a bit of a sloppy question if we don't state that x must be an integer. I don't think the official test would ask a question like this without saying that x is an integer.

IanStewart · Answer

The equation in the stem is a bit of a famous one in math, and it has three solutions in total, the two positive integer solutions 2 and 4, and one negative non-integer solution, which is close to -0.77. It's an example of something called a "transcendental equation" and those equations typically have some solutions which can't be expressed neatly (using integers and roots, say), which is the case with the negative solution here.

But as Bunuel points out above, this is an official question, and this is its wording. There's nothing wrong with the wording since there's no need to identify the non-integer solutions to any of the given equations. If you just combine the equations immediately (rather than solve them individually to see where the solutions overlap), only one integer solution is possible.

ScottTargetTestPrep · Answer

Solution:
 
We need to determine the value of x given that x^2 = 2^x. Notice that the equation has at least two solutions, namely 2 and 4.

Statement One Alone:
 
Solving the equation, we have:

2x = x^3 / 8

16x = x^3

x^3 - 16x = 0

x(x^2 - 16) = 0

x(x - 4)(x + 4) = 0

x = 0 or x = 4 or x = -4

Since only 4 satisfies the equation x^2 = 2^x, then x must be 4. Statement one alone is sufficient.

Statement Two Alone:

Let’s substitute x = 2^(x - 2) in the equation x^2 = 2^x:

^2 = 2^x

2^(2x - 4) = 2^x

2x - 4 = x

x = 4

Statement two alone is sufficient.

Answer: D

aditisethi6 · Answer

Statement (1): x on LHS cancels with RHS and we are left with 2 = x^2/2^3 i.e. 2^x = 2^4;

Hence x = 4

Crytiocanalyst · Answer

(1) 2x=(x2)32x=(x2)3
x∗(x28−2)=0x∗(x28−2)=0
we get x as 0, 4 or -4..
when we substitute value in x2=2xx2=2x and the only value that is in accordance with given data and statement 1 
sufficient

(2) x=2x−2x=2x−2
or x=2x−2x=2x−2...
which gives us 4x-x^2 = 4 or 0 , 0 is eleminated and the only value that stands out is 4 
hence each statement is individually sufficient hence IMO D

amitupreti · Answer

x =2 is also the solution, but neither St.1 not St.2 give us x=2 as the solution. Shouldn't the answer be E ?

BrentGMATPrepNow · Answer

Given :  x^2 = 2^x

Strategy: There aren't many solutions to this given equation, so let's spend a little bit of time up front listing all possible cases.

It turns out there are only two possible cases: 
Case i:  x = 2 
Case ii:  x = 4 
In other words, the given information is telling us that   x is either 2 or 4  . 
With this in mind, it'll be super easy to check each statement.

Target question :    What is the value of x?

Statement 1 :  2x = (\frac{x}{2})^3   
Rather than solve this equation, we can just test the two possible x-values.

Test  x =2 , to get:  2(2) = (\frac{2}{2})^3 , which we can simplify to get:  4 = 1 . FALSE. 
So,  x 
eq 2 , which means it must be the case that   x = 4  
Since we can answer the  target question  with certainty, statement 1 is SUFFICIENT

Statement 2 :  x = 2^{x-2}  
Once again, we'll test the two possible x-values.
Test  x =2 , to get:  2 = 2^{2-2} , which we can simplify to get:  2 = 1 . FALSE. 
So,  x 
eq 2 , which means it must be the case that   x = 4  
Since we can answer the  target question  with certainty, statement 2 is SUFFICIENT

Answer: D

Cheers,
Brent

Elite097 · Answer

avigutman  how do we know here that x cannot take fractional or irrational values? We are not told x is integer. Not clear on existing explanations

IanStewart  did not get your whole of last explanation where you defend that the wording is right. I do not think it is right

IanStewart · Answer

x actually can take an irrational value in the equation in the stem, and also in the equation in Statement 2, as WheatyPie and I mentioned above. x just can't take the same irrational value in both equations, so fortunately we don't need to think about the non-integer solutions here (the irrational solutions to those two equations are impossible to find using GMAT-level math, except by approximation, so the GMAT could never ask a question where you needed to find them).

When you solve the two equations together (the one in the stem, and the one in Statement 2) you're finding every legitimate value of x that works in both equations. So if, say, we rewrite the equation in the stem as x^2 = 4(2^(x-2)) and then, using Statement 2, replace 2^(x-2) with x, we find x^2 = 4x, and since x is nonzero, x = 4. So we only get one solution, which turns out to be an integer.

Elite097 · Answer

I donot get it. If the irrational value works in both question stem and statement 2 why does it not work when those two are used together  IanStewart ??

IanStewart · Answer

The irrational value that works in the stem is not the same as the irrational value that works in Statement 2.

If x^2 = 2^x, what is the value of x ?

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