Find explicitly the positive solutions of the equation $2^x=x^2$
I noticed that $x=2$ and $x=4$ are roots of the equation.
How can I prove that they are the only positive ones? Thanks in advance
Asked
Active
Viewed 65 times
2
-
1How familiar are you with derivatives? – Arthur Apr 21 '18 at 14:56
-
I am quite familiar with them – Lorenzo B. Apr 21 '18 at 14:58
-
you have $2^x =e^{loga}$ and you have to study the functions $2^{x}-x^2$ – Bernstein Apr 21 '18 at 14:58
3 Answers
3
Let $f(x)=2^x-x^2$. Then by implicit differentiation $$f'(x)=2^x\ln2-2x,\quad f''(x)=2^x\ln^22-2$$ Now solving $f''(x)=0$ gives $$x_0=1-\frac{\ln\ln^22}{\ln2}$$ which is the global minimum of $f'(x)$.
Since $$f'(0)=\ln2>0,\quad f'(x_0)<0,\quad f'(4)=16\ln2-8>0$$ and $f'$ is continuous in $\mathbb{R}$, by the Intermediate Value Theorem there are exactly two solutions to $f'(x)=0$.
Can you continue?
ə̷̶̸͇̘̜́̍͗̂̄︣͟
- 27,005
-
1
-
@Arthur Finding $\frac d{dx}(2^x)$ requires taking for the log and using the chain rule – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Apr 21 '18 at 15:41
-
You could look at it that way. Or you could just start with the definition of $2^x$, namely $2^x=e^{x\log(2)}$. (Yes, that's close to equivalent, but it explains how Arthur and I didn't see anything "implicit"...) – David C. Ullrich Apr 21 '18 at 15:44
-
That's still not what I consider implicit differentiation. Implicit differentiation is when you have some curve defined by an expression which isn't necessarily a function (like the circle $x^2+y^2=1$), and then you pretend $y$ is a function (a so-called implicit function) of $x$ and differentiate both sides of that expression to figure out how steep the curve is at a given point. Implicit isn't about what tools you use, but whether what you differentiate is actually a function, or just something you're pretending to be a function. – Arthur Apr 21 '18 at 15:50
1
They're not the only solutions, if we also allow negative nummbers:. Since $2^0>0^2$ and $2^{-1}<(-1)^2$ there exists a solution $x$ with $-1<x<0$.
Those three are the only real solutions. Let $f(x)=2^x-x^2$. Calculate the third derivative and show that $f'''(x)>0$ for every $x$. If $f$ had four zeroes the Mean Value Theorem would show that $f'$ had three zeroes, hence $f''$ would have two zeroes, hence $f'''$ would have a zero.
So since there are only three real solutions and one is negative there are only two positive solutions.
David C. Ullrich
- 89,985
-
2
-
I am sorry I didn't make myself clear, I was trying to prove they are the only positive solutions – Lorenzo B. Apr 21 '18 at 15:13
-
@Lorenzo Sorry, I missed the word "positive". If you know there are exactly three real solutions, you know there are at least two positive solutions and at least one negative solution it follows trivially that there are exactly two positive solutions. – David C. Ullrich Apr 21 '18 at 15:18
-
how does the mean value theorem explain that $f''$ has 2 zeros and $f'$ three zeros? – The Integrator Apr 21 '18 at 15:19
-
@pranavB23 Do you know the mean value theorem? Suppose $f$ has four zeroes $a<b<c<d$. MVT shows that $f'$ has a zero in $(a,b)$, that $f'$ has a zero in $(b,c)$ and that $f'$ has a zero in $(c,d)$. – David C. Ullrich Apr 21 '18 at 15:22
-
@DavidC.Ullrich yeah i know the mean value theorem but im unclear on how it relates to the second and third derivatives – The Integrator Apr 21 '18 at 15:25
-
1@pranavB23 In my last comment I gave a detailed proof that if $f$ has four zeroes then $f'$ has at least three zeroes. Exactly the same argument shows that $f''$ has at least two zeroes. This is because $f''$ is the derivative of $f'$. – David C. Ullrich Apr 21 '18 at 15:39
-
0
For positive $x$, we have $2^x=x^2\iff a^x=x$, where $a=\sqrt2$. Since $0\lt a\not=1$, the exponential curve $y=a^x$ is concave up, hence any line intersects it at most twice. The line $y=x$ does so at $x=2$ and $x=4$. So those are the only positive solutions of $2^x=x^2$. (As David Ullrich notes, there is also a negative solution; it corresponds to the interesection of the exponential curve $y=a^x$ and the line $y=-x$.)
Barry Cipra
- 79,832