# Math Help - 2^x = x^2

1. ## 2^x = x^2

The instruction is to "use your calculator" to solve 2x = x2, as opposed to solving analytically (leaving logs unreduced etc). Judging from the answer, which includes 2 and some decimal, this might be a quadratic in form somehow perhaps. Anyway, after hours of mulling it I can't seem to find the correct approach. It's easy to get to .3466 = ln(x)/x. Can't get passed that. I also considered (x - 2sqrt(x)) * (x + 2sqrt(x)) but I couldn't get anywhere with that.

2. ## Re: 2^x = x^2

Did you consider doing what you were told to do? That is, "use your calculator", as opposed to "solving analytically". This cannot be solved "analytically", using elementary functions, which is what you are trying to do. (The answer is NOT "2 and some decimal". There is, in fact, two simple solutions you could get by 'inspection', one that's a bit more complicated. But, ignoring that...)

Instead, consider graphing 2^x and x^2 on your (graphing) calculator and "zooming" in on the point at which they cross.

Or do this. Let f(x)= 2^x- x^2 which will be equal to 0 if x satisfies 2^x= x^2. If x= -1, f(-1)= 2^-1- (-1)^2= 1/2- 1= -1/2. Since f(1) is positive, f(-1) is negative, and f is continuous, there must be a solution somewhere between -1 and 1. Where? Don't know! But since it is easy to divide by 2, lets look halfway between: (-1+ 1)/2= 0/2= 0. If x= 0, f(0)= 2^0- 0^2= 1. That is positive so there must be a solution somewhere between -1 and 0. Again, try halfway between (just because it is easy). (-1+ 0)/2= -1/2. (2^{-/2}- (-1/2)^2= [tex]1/\sqrt{2}- 1/4= 0.4571, approximately. That is still positive (but close to 0) so there must be a solution between -1 and -1/2. Again, half way between is -3/4= -.75. f(-3/4)= 2^{-3/4}- (-3/4)^2= 0.5946- 0.5625= 0.0321. Again, that is positive so now we know there is a solution between -.75 and -1. Continue until you have enough accuracy.

(There are two positive solutions but those are integers so I would not use this method!)

The simplest way to solve this equation is what I first suggested: graph $y= 2^x$ and $y= x^2$ and zoom in on the intersections.

3. ## Re: 2^x = x^2

Well the answer is '2 and some decimal', namely -.76666. I appreciate the approach based on the continuity of f (and the associated theorem). I am a bit surprised that there is no way to calculate the answer, other than through brute force. It never occurred to me that they meant use the graphing function of the calculator rather than computing logs using ln. Thanks!

4. ## Re: 2^x = x^2

Hi,
I agree with HallsOfIvy; you really have to compute numerically (graphic calculator can do this). However, the reason I'm answering is to point out a flaw in your attempt to solve analytically. Namely:
$2^x=x^2\text{ , then }x\text{ln}2=\2\text{ln}x$
This is true only if x>0! So even if somehow you were to solve the 2nd equation, you'd miss the root -.7666.
Same objection to your second "simplification". Moral: in solving equations, be aware that a step may not be reversible; check for this possibility.

5. ## Re: 2^x = x^2

Originally Posted by kjtruitt
Well the answer is '2 and some decimal', namely -.76666.
I misunderstood what you meant. I though '2 and some decimal' meant '2 point something'. Yes, 2 and about -.766 are two solutions. But you have missed another obvious one: x= 4. $2^4= 16= 4^2$.

I appreciate the approach based on the continuity of f (and the associated theorem). I am a bit surprised that there is no way to calculate the answer, other than through brute force. It never occurred to me that they meant use the graphing function of the calculator rather than computing logs using ln. Thanks!

6. ## Re: 2^x = x^2

Hi !

There are two obvious roots : x=2 and x=4.
The third root involves a special function : the Lambert W function. Of course the wording of the question specifies that you must not use this function. So, you have to compute an approximate value of the third root thanks to a numerical computation method
solve 2^x=x^2 - Wolfram|Alpha

7. ## Re: 2^x = x^2

The focus right now is just learning everything about exponential and logarithmic functions. Even if I could spot the obvious ones it seems that they'd want me to perform the right manipulations of the equation--but I'm told here there really aren't any (other than this Lambert W function). I appreciate your comments