Prove that $e^{2x} = 3x$ has no solution

Question

How can I prove that the equation:

$ e^{2x} = 3x $

has no solution?

I thought to use Rolle's theorem, but I didn't managed to do so, could you help?

Answer 1

All of the roots must be positive, because for $x<0$ the RHS is negative, but the LHS is always positive (and $x=0$ is not a root, of course). So from now on, let's assume that $x>0$.

Let $$e_n(x):= \sum_{i=0}^{n} \frac{x^i}{i!}$$ It's easy to see that $$e_{n+1}-e_{n} = \frac{x^{n+1}}{(n+1)!} \geqslant 0$$ So $(e_n(x))_{n \in \mathbb{N}}$ is an increasing sequence. And because $$e^x=\sum_{n \in \mathbb{N}} \frac{x^n}{n!}=\lim_{n} e_n(x)$$ We have that $$e^{2x} \geqslant e_{2}(2x)=1+2x+2x^2$$ $$e^{2x} -3x \geqslant 1 -x + 2x^2$$ But the $1-x+2x^2$ is always positive, which implies that: $$e^{2x}-3x>0$$ So $e^{2x}=3x$ has no real root.

Answer 2

When $x>0$ (why worry about when $x\le0$?) then $$e^{2x}>1+2x+2x^2$$ and $$e^{2x}-4x>1-2x+2x^2=(1-x)^2+x^2.$$ You only wanted $e^{2x}>3x$?

Answer 3

$f(x)=e^{2x}$ is clearly a convex function, and the slope of the tangent line to the graph of $f$ equals $3$ iff $x=\frac{1}{2}\log\frac{3}{2}$. By Jensen's inequality it follows that

$$ \forall x\in\mathbb{R},\qquad f(x)=\color{red}{e^{2x}} \geq 3\left(x-\tfrac{1}{2}\log\tfrac{3}{2}\right)+\tfrac{3}{2}\color{red}{\geq 3x+\tfrac{8}{9}} $$ and $f(x)=3x$ is clearly impossible.

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Alternative formulation: $g(x)=e^{2x}-3x$ is trivially convex (it is enough to compute $g''$). The only root of $g'$ is at $x=\tfrac{1}{2}\log\tfrac{3}{2}$, where $g(x)>0$. It follows that $g(x)>0$ for any $x\in\mathbb{R}$.

Answer 4

Let $$f(x)=e^{2x}-3x$$ Then $$f'(x)=2e^{2x}-3=0$$ for stationary points. So $$e^{2x}=\frac32\implies 2x=\ln\frac32\implies x=\frac12\ln\frac32=\ln\sqrt{\frac32}$$ Since $$f''(x)=4e^{2x}$$ at this $x$, the second derivative is $$4\cdot \frac32=6>0$$ so it is a minimum. There are no other stationary points.

Now notice that at this $x$, $f$ is equal to $\frac32-3\ln\sqrt{\frac32}>0$, meaning that $f$ is always above the $x$-axis.

Thus $e^{2x}=3x$ has no solutions.

Answer 5

Rewriting $e^{2x} = 3x$ as $$-2x e^{-2x} = -\frac{2}{3}, \tag1$$ we see (1) is exactly in the form for the defining equation for the Lambert W function, $\text{W} (x)$. On solving for $x$ in terms of this function we have $$-2x = \text{W}_\nu \left (-\frac{2}{3} \right ),$$ or $$x = -\frac{1}{2} \text{W}_\nu \left (-\frac{2}{3} \right ).$$ Here $\nu$ denotes the branches for the Lambert W function. If $\text{W} (x)$ is to be real we require its argument to be greater than or equal to $-1/e$ (in which case the branches $\nu = -1,0$ will be selected when $-1/e \leqslant x < 0$ and $\nu = 0$ when $x \geqslant 0$). As $-2/3 < -1/e$ we conclude the equation are no real solutions.