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The question in the title is equivalent to find the number of the zeros of the function $$f(x)=x^{12}-2^x$$

Geometrically, it is not hard to determine that there is one intersect in the second quadrant. And when $x>0$, $x^{12}=2^x$ is equivalent to $\log x=\frac{\log 2}{12}x$. There are two intersects since $\frac{\log 2}{12}<e$.

Is there other quicker way to show this?

Edit: This question is motivated from a GRE math subject test problem which is a multiple-choice one(A. None B. One C. Two D.Three E. Four). Usually, the ability for a student to solve such problem as quickly as possible may be valuable at least for this kind of test. In this particular case, geometric intuition may be misleading if one simply sketch the curve of two functions to find the possible intersect.

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This may or may not be helfpul: – JavaMan Jun 8 '11 at 23:08
Yeah, you can get $ (1/x) \ln (1/x) = -(\ln 2) / 12$ so you can conclude $ x = \exp \left[ - W(-\frac{\ln 2}{12}) \right] $. Since the argument is in the interval $ (-1/e, 0) $ you can conclude the W function is two-valued and hence there are two solutions to the original problem. The only issue here is we're outsourcing the proof to facts given on Wikipedia, so not fully explicit as it could be. Don't know if that's satisfactory for you or not. – anon Jun 8 '11 at 23:24
@DJC/@anon: +1,nice comment about the W function! – Jack Jun 8 '11 at 23:29
up vote 6 down vote accepted

We need to remember that this was a multiple choice test (ugh!) so to talk about strategies we need to know the choices that were offered. There is not enough room in the little bubbles that students are supposed to black in for even a short proof.

A rough knowledge of the graphs was probably enough. Even the most primitive sketch, or mental image of a sketch, shows that there is precisely one intersection point with $x$ negative.

For $x \ge 0$, $2^x$ starts above $x^{12}$. And $2^x$ is grossly below $x^{12}$ at $x=2$. And $2^x$ is ultimately going to "win" by standard properties of the exponential.

So there are at least $3$ intersection points. I imagine that the choices were something like $1$, $2$, $3$, $4$. So we probably only need to eliminate $4$. This can be done by noting that the number of intersection points with $x \ge 0$ is probably even ($2^x$ starts above, ends up above, who would ever think of the conceivability of tangency).

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Choices added. Thanks:-) – Jack Jun 9 '11 at 2:13

If you are solving a multiple choice test like GRE you really need fast intuitive, but certain, thinking. I tried to put myself in this rushed set of mind when I read your question and thought this way: think of $x^{12}$ as something like $x^2$ but growing faster, think of $2^x$ as $e^x$ similarly, sketch both functions.

It is immediate to see an intersection point for $x<0$ and another for $0<x<b$, for some positive $b$ since the exponential grows slower for small $x$ for a while, as the sketched graph suggests. So the answer is at least $2$. In fact it is $3$ because after the second intersection point you clearly see the graph of $x^{12}$ over $2^x$ but you should notice that $a^x\gg x^n$ at $+\infty$, and therefore the exponential must take over and have a third intersection point at a really big value of positive $x$. Once this happens the exponential function is growing so fast that a potential function cannot catch up so there are no further intersections.

(To quickly see that $a^x\gg x^n$ at $+\infty$ just calculate $\lim_{x\to\infty}\frac{a^x}{x^n}=+\infty$ using L'Hopital's rule or Taylor expanding the numerator whose terms are of the form $\log^m(a)a^m/m!$).

More rigorously, maybe you can find a way to study the signs of $g(x)=x^{12}-2^x$ using derivatives and monotony. There are 4 intervals giving signs + - + - resulting in 3 points of intersection by the intermediate value theorem. These intervals are straightforwardly seen as reasoned above just sketching the function and taking into account the behavior for big values of $x$. To be sure that there is no other change of sign you must prove that $g'$ is monotone after the third point of intersection. Just after this last point, the graph of $2^x$ can easily be seen over $x^{12}$ and both subfunctions are monotone along with their derivatives: since $2^x>x^{12}\Rightarrow \log(2)2^x>12x^{11}$ which means $g'(x)=12x^{11}-\log(2)2^x$ is indeed monotone afterwards and therefore there is no fourth intersection.

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This is not really an independent answer, more a gathering in one place of what others have said.

It's easy to see there's a solution between $-1$ and $0$ (since $x^{12}-2^x$ is positive at $-1$ and negative at $0$), and another solution between $1$ and $2$ (again, there's a change of sign), and another solution bigger than $2$ (by Theo Buehler's comment on Joseph Malkevitch's answer). Then there are any number of ways to show that there are no solutions other than these three.

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Indeed, just use the derivative of $g(x)=x^{12}-2^x$ and study monotony on the interval between the second solution and $+\infty$ – Javier Álvarez Jun 9 '11 at 2:41
@Javier: For studying the monotony of $g(x)$, one may needs to determine the zeros of $g'(x)$ which may not be a trivial job here. – Jack Jun 9 '11 at 2:48
@Jack: You are right. Nevertheless it is useful to rest assured there is no 4th intersection point rigorously which is why I did my previous comment referring to the last sentence of Gerry Myerson's answer "there are a number of ways to show that there"[is no 4th solution]. I have added a small explanation at my posted answes. – Javier Álvarez Jun 9 '11 at 3:44
@Javier: Hmm, indeed, if one needs to "prove" it, $g'(x)$ is needed here. – Jack Jun 9 '11 at 3:48

From Maple... $$ \Biggl[\frac{-12\mathrm{LambertW} \biggl(\frac{\operatorname{ln} (2)}{12}\biggr)}{\operatorname{ln} (2)},\frac{-12\mathrm{LambertW} \biggl(\frac{-\operatorname{ln} (2)}{12}\biggr)}{\operatorname{ln} (2)}, \frac{-12\mathrm{LambertW} \biggl(-1,\frac{-\operatorname{ln} (2)}{12}\biggr)}{\operatorname{ln} (2)}\Biggr] $$ $=[-0.9467803304,1.063346831,74.66932553]$ plus many complex solutions.

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Graph the function $f(x) = x^{12} 2^{-x}$ for $x\ge 0$. Then $$f'(x) = x^{11}\left(12 - x\log(2)\right)2^{-x},$$ once you clean things up. For $x > 0$, we see the sign of $f'$ is entirely determined by $12 - x\log(2)$.

Now draw the sign chart for $f'$ on $[0,\infty)$. You see that $f$ is increasing before $12/\log(2)$ and decreasing thereafter. Since $f(0) = 0$, $f > 1$ at its maximum, and $f(x) \rightarrow 0$ as $x\to\infty$, the graph of $f$ crosses the vertical line $y = 1$ exactly twice.

You have exactly two zeroes.

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The simple kind of graphing skills one develops in analytical geometry/pre-calculus would suggest the answer would be two. One can try Wolfram Alpha to see what happens:

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@Joseph: I've also tried MATLAB to get a geometric intuition. However, I think in the first quadrant, there are two roots. As GEdgar's answer shows, another positive root is not that obvious in you picture. – Jack Jun 8 '11 at 23:47
@Jack: There must be a third one, as $2^x \gt x^n$ for $x \gg 0$. Since for $x = 2$ we have $4 = 2^2 \lt 2^{12}$ a third one must exist by the intermediate value theorem. – t.b. Jun 8 '11 at 23:51
@Theo: How can one get the estimate that $2^x>x^n$ for $x\gg0$? – Jack Jun 9 '11 at 2:09
@Jack: Let me do it for $e^{x} \gt x^n$. We know that $e^{x} = \sum_{n=0}^\infty \frac{x^n}{n!} \gt \frac{x^{n+1}}{(n+1)!}$ (the $\gt$ follows from $x \gt 0$). So for $x \gt (n+1)!$ we have $e^{x} \gt \frac{x^{n+1}}{(n+1)!} = \frac{x}{(n+1)!} \cdot x^n \gt 1 \cdot x^n = x^n$ as we wanted. Similarly $e^{cx} \gt \frac{(cx)^{n+1}}{(n+1)!}$ if $c \gt 0$. Can you get it from here? (you're interested in $c = \log{2}$). – t.b. Jun 9 '11 at 2:21
@Jack: That was the exact way. What you really need to remember and never forget: 1. polynomials of higher degree grow eventually faster than those of lower degree; 2. the exponential function grows faster than any polynomial. That's why $2^x \gt x^n$ for $x$ large enough. (one millisecond of thought if you know 2.) – t.b. Jun 9 '11 at 2:35

taking the function $g(x) = \log(x) -x \log(2) /12$ and observe where it is monoton. There it can have at most one zero, i.e. in the interval $( - \infty , 0)$, $(0, 12/ \log(2))$ and $(12 / \log(2), +\infty)$. Then a case by case analysis should solve the problem.

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Use \log instead of log – t.b. Jun 8 '11 at 23:44

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