# Solve $e^x = 2x$ with algebra

I’m trying to solve $$e^x = 2x.$$ I have already understood that there isn’t any solution when I plot both functions, but I would know how to get this conclusion by trying to solve it algebrically. I got owned with: $$\ln(e^x) = \ln(2x)$$ $$x\ln(e) = \ln(2x)$$ $$x = \ln(2x)$$ Then I don’t know how to continue. I tried to pass it in WolframAlpha and other computation but the only thing it tell me that there is no solution.

• it hasn't solution 'cuz their plots clearly indicate there is no intersection wolframalpha.com/input/?i=plot+%7Be%5Ex%2C2x%7D Sep 25, 2021 at 19:13
• maybe by contradiction Sep 25, 2021 at 19:14
• You can use $e^x \geq 1 + x + \frac12x^2$ for $x \geq 0$, and $e^x > 0$ everywhere (in particular, for $x < 0$). Sep 25, 2021 at 19:17
• Show that $2x < 1 + x + \frac12x^2$ everywhere, and $2x < 0$ for $x < 0$. Sep 25, 2021 at 19:19
• Oh, I understand your confusion now. Those are the first three terms of the Taylor series expansion of $e^x$. For $x \geq 0$, all the terms are non-negative (and strictly positive for $x > 0$). Sep 25, 2021 at 19:24

We see that $$e^x=2x\implies x>0$$. Because, $$e^x>0,\thinspace \forall x\in\mathbb R$$.

Therefore, we have

\begin{align}&e^x=2x,\thinspace x\in\mathbb R\\ \implies &x=\ln 2+\ln x, \thinspace x>0\\ \implies &e^u-u=\ln 2,\thinspace u=\ln x \\ \implies &\ln2+u≥u+1;\thinspace e^u≥u+1\\ \implies &\ln 2≥1 \\ \implies &e≤2 \\ &\text{A contradiction.}\end{align}

• There are actually complex solutions using a product log function. Sep 25, 2021 at 19:30
• @TymaGaidash Yes, Lambert-W works. Sep 25, 2021 at 19:32
• Where is your second implication coming from? Adding $u$ to both sides of the second equality doesn't get you there because $\ln 2 \lt 1$. Sep 25, 2021 at 20:23
• @Robert $\ln 2+\ln x=x\implies x-\ln x=\ln 2 \implies e^u-u=\ln 2$. Where is the error? Sep 25, 2021 at 20:26
• You've changed your second implication. How do you know that $e^u \geq u+1$? Doesn't that require use of the Taylor series? Sep 25, 2021 at 22:52

As I indicate in the comments, you can use the bounds

$$e^x > \begin{cases} 0 & x \leq 0 \\ 1+x+\frac12x^2 & x > 0 \end{cases}$$

and

$$2x \leq \begin{cases} 0 & x \leq 0 \\ 1+x+\frac12x^2 & x > 0 \end{cases}$$

The expression $$1+x+\frac12x^2$$ comes from the first three terms of the Taylor series expansion for $$e^x$$; all terms are non-negative for $$x \geq 0$$.

• The use of a Taylor series makes it seem to me like not an algebraic proof, as was requested. Sep 25, 2021 at 20:21
• @RobertShore: One must have some definition of $e^x$. It could just be the Taylor series itself, in which case there is no more direct route. (I would say that is a very common definition, in fact—perhaps the most common.) Another definition could be that it is the unique solution to $\frac{dy}{dx} = y, y(0) = 1$, in which case I believe there is no point to a "purely algebraic" solution, whatever that might be. In some sense, the involvement of $e^x$ already implies a partly non-algebraic solution. Sep 25, 2021 at 21:26
• Or, alternatively, you could define $e^x$ as $\lim_n\to\infty \left(1+\frac1n\right)^n$. I don't happen to prefer this one, but if you use it, you can apply a generalized binomial theorem and recover the Taylor series. Sep 26, 2021 at 20:02

Let $$f(x)=e^x-2x$$ then $$f'(x)=e^x-2$$ and it annulates only in $$x_0=\ln(2)$$.

Therefore $$f\searrow$$ then $$\nearrow$$ with a minimum in $$x_0$$.

The minimum $$f(x_0)=e^{\ln(2)}-2\ln(2)=2-2\ln(2)>0$$ so the function has no zeroes.

Making the problem more general, consider that you look for the zero of function $$f(x)=e^x-kx$$ the solution of which being given in terms of Lambert function $$x=-W\left(-\frac{1}{k}\right)$$ This solution is real only for $$k \geq e$$.

So, if $$k, assuming that you do not want to use Lambert function, let $$xa+b i$$ to make $$e^x-k x=\Big[e^a \cos (b)-a k\Big]+i \big[e^a \sin (b)-b k\Big]$$ Cancelling the real part, we have $$e^a \cos (b)-a k=0 \implies b=\cos ^{-1}\left(a e^{-a} k\right)$$ Plugging it in the imaginary part leads to $$g(a)=k \cos ^{-1}\left(a e^{-a} k\right)-e^a \sqrt{1-a^2 e^{-2 a} k^2}=0$$ which is difficult to solve using a numerical method such as Newton.

We can approximate the solution using a series expansion around $$a=0$$

$$g(a)=\left(\frac{\pi k}{2}-1\right)- \left(k^2+1\right)a+\frac{3k^2-1}{2} a^2 -\frac{\left(k^2+6\right) k^2+1}{6} a^3 +\frac{15 k^4+10 k^2-1}{24} a^4 +O\left(a^5\right)$$ Using series reversion $$a=t+\frac{\left(3 k^2-1\right) t^2}{2 \left(k^2+1\right)}+\frac{\left(-k^6+20 k^4-25 k^2+2\right) t^3}{6 \left(k^2+1\right)^2}+\frac{\left(-15 k^8+245 k^6-511 k^4+183 k^2-6\right) t^4}{24 \left(k^2+1\right)^3}+O\left(t^5\right)$$ where $$t=\frac{\pi k-2}{2 \left(k^2+1\right)}$$.

Using it for $$k=2$$ as in your case, this would give as an estimate $$a_0=\frac{-30391+23984 \pi +6144 \pi ^2-176 \pi ^3+439 \pi ^4}{187500}=0.762140\cdots$$ which would give $$b_0=0.779407$$.

If you want to polish the root of $$g(a)=0$$, using Newton method, the iterates will be $$\left( \begin{array}{cc} n & a_n \\ 0 & 0.762140 \\ 1 & 0.793950 \\ 2 & 0.794024 \end{array} \right)$$ This will give $$b=0.770112$$.