# Interesting GRE problem

I found this in a practice GRE problem. I thought I would have a crack at it after being spoiled by the answer

At how many points in the xy-plane do the graphs of $y = x^{12}$ and $y = 2^x$ intersect?

So I thought about doing something what most people would have done, solving for the intersection, $x^{12} = 2^x$, it became utterly hopeless.

Then I thought about using the Intermediate Value Theorem, that is

$f=x^{12} - 2^x = 0$

I suspect for $x<0$, $x^{12} > 2^x$, so $f>0$

For $x=0$, $f < 0$. So by IVT, there is a root somewhere between $(-\infty,0)$

For $x>0$, $x^{12} > 2^x$, so $f>0$. So by IVT, there is another root at $(0,\infty)$

So counting, I should get 2 roots, another therefore 2 points. But the actual answer was 3. So I am inclined to believe I overlooked something very important

Note: The GRE forbids technology assistance.

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Set them equal and take the log of both sides. Then you probably can pick out a few by inspection. This is supposed to be a precalc problem and they don't know much. If this is on the GRE, I wouldn't be surprised if the people who cooked up the problem made a mistake. – toypajme Dec 21 '12 at 19:55
@toypajme, that was my first attempt. It didn't seem like it could lead anywhere. So I tried something else – Hawk Dec 21 '12 at 19:58
Exactly 3 points. – copper.hat Dec 21 '12 at 19:59
Hint: Look at the plot of the curves on top of each other regarding your first suspicion. – Amzoti Dec 21 '12 at 20:02
The argument for $x\gt 0$ was not right. In the long run $e^x\gt x^{12}$. But $e^2\lt 2^{12}$. So there is a root between $0$ and $2$, and a root somewhere beyond $2$. – André Nicolas Dec 21 '12 at 20:02

$x^{12} = 2^x$ (for $x$ real) is equivalent to: either $x = 2^{x/12}$ or $-x = 2^{x/12}$. Since $2^{x/12}$ is convex, its graph intersects a straight line in $0$, $1$ or $2$ points.

$-x$ is decreasing while $2^{x/12}$ is increasing, and $-x > 1 > 2^{x/12}$ for $x < -1$ while $-x < 2^{x/12}$ for $x > 0$, so there is exactly one real solution of $-x = 2^{x/12}$ and it is in the interval $-1 \le x \le 0$.

$x < 0 < 2^{x/12}$ for $x < 0$, $x < 2^{x/12}$ for sufficiently large $x$, while $x > 2^{x/12}$ at $x=2$, so there are two real solutions of $x = 2^{x/12}$, one with $0 < x < 2$ and one in $2 < x < \infty$.

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Oh this is interesting take I like – Hawk Dec 21 '12 at 20:34

I suspect for $x<0$, $x^{12} > 2^x$, so $f>0$

Obviously false for negative $x$ close to 0.

For $x>0$, $x^{12} > 2^x$

Obviously false for large positive $x$.

But really, all you have to do is draw a graph. Go away and draw one, and then come back and tell us about it :-)

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You can't bring graphing devices into a GRE. We are suppose to do this without the assistance of technology I believe – Hawk Dec 21 '12 at 20:08
Even if I were to draw a sketch, I could only hit two obvious points. The third one is too subtle for me considering it lands on $x \approx 74...$ – Hawk Dec 21 '12 at 20:10
@sizz, by "draw" he means sketch. Also, the third point is known because $2^x$ grows quicker than $x^{12}$. – picakhu Dec 21 '12 at 20:11
@sizz, you simply ignored my first two points. – TonyK Dec 21 '12 at 20:28

I thought of it this way; near the origin, $2^x$ is relatively flat, while $x^{12}$ points up sharply. This gives two points of intersection. But, eventually, exponentials outgrow any polynomial, so there must be another point of intersection where $2^x$ outgrows $x^{12}$.

Actually, I didn't think of the third root at the time either and put down 2 as my answer on the practice exam. But hey, only 12 percent of test takers get it right!

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This is how I did it. There seem to be a few problems like this on each GRE, where if you try do things rigorously it's impossible, but intuitively it's obvious once you see the point of the problem. – Potato Sep 25 '14 at 1:04

First thing to observe is that both the functions are continuous and the graphs are "smooth" curves.

There is one and only one point of intersection in $(- \infty, 0)$ as $x^{12}$ is increasing & $2^x$ is decreasing as $x \rightarrow- \infty$ and $2^0 > 0^{12}$.

Now observe that when $x \rightarrow \infty,$ both the functions increase and $2^x$ dominates $x^{12}$ eventually. Since $2^{12} > 2^2$, it must be the case that the graphs intersect at some point in $(2, \infty)$. This is the only point of intersection in that interval.

Also, $2^0 > 0^{12}$ and $2^{12} > 2^2$ tells us that there is a point of intersection in $(0,2)$ and there is exactly one such intersection.

So altogether there are exactly 3 points of intersection.

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