How to find the solutions for the equation $2^x = 4x$?
How do I solve the equation $2^x = 4x$? So, the obvious answer by inspection is $x=4$, but I need an analytical solution to this, something rigorous pls.
$\endgroup$ 64 Answers
$\begingroup$Let $f(x)=2^x-4x$.
Thus, $$f''(x)=2^x\ln^22>0,$$ which says that $f$ is a convex function, which gives that our equation has two real roots maximum.
But $4$ and $0<x_1<1$ are roots.
The value of $x_1$ we can not get by elementary functions.
$\endgroup$ $\begingroup$$$2^x=4x$$ $e^{x\ln(2)}=4x\quad\to\quad xe^{-x\ln(2)}=\frac14\quad\to\quad \left(-x\ln(2)\right)e^{\left(-x\ln(2)\right)}= -\frac{\ln(2)}{4}$
The general solution of this kind of equation cannot be expressed with a finite number of elementary functions. We need infinite series or a special function : The Lambert's W function,
$$-x\ln(2)=W\left(-\frac{\ln(2)}{4} \right)$$ The result is : $$x=-\frac{1}{\ln(2)}W\left(-\frac{\ln(2)}{4} \right)$$ The $W(X)$ function is a multi-valuated function. In the real range $-e^{-1}<X<0$ they are two real roots : $$\begin{cases}W_{-1}\left(-\frac{\ln(2)}{4} \right)=-4\ln(2) \quad\to\quad x=4\\W_0\left(-\frac{\ln(2)}{4} \right)\simeq-0.214811 \quad\to\quad x \simeq 0.3099069 \end{cases}$$ The second root is not rational. An approximate is computed thanks to any mathematical software where the Lambert function is implemented. Or alternatively with recursive numerical calculus (Newton-Raphson method for example).
$\endgroup$ $\begingroup$A bit of detail:
$ f''(x) \gt 0$, $f= 2^x-4x$ convex, does not , by itself, imply that $f(x) = 0$ has $2$ distinct zeroes.
1)$y= x^2 +1$, convex, has no zeroes ,
2) $y= x^2,$ $x_1=x_2 =0$, double zero.
But since $f(x) = 2^x -4x$ is convex, and:
positive at $x= 0,$
negative at $x=1$ ,
positive for $x \gt 4,$
we conclude that $f(x)$ has two distinct roots:
$x_1 = 4$, by inspection, and $0 \lt x_2\lt 1$.
$\endgroup$ 2 $\begingroup$For a rigorous proof you have to study the function $f(x)=2^x-4x$ showing analitically that it can't have more than two roots.
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