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.
 A: We see that $e^x=2x\implies x>0$. Because, $e^x>0,\thinspace \forall x\in\mathbb R$.
Therefore, we have
$$\begin{align}&e^x=2x,\thinspace x\in\mathbb R\\
\implies &x=\ln 2+\ln x, \thinspace x>0\\
\implies &e^u-u=\ln 2,\thinspace u=\ln x \\ 
\implies &\ln2+u≥u+1;\thinspace e^u≥u+1\\ 
\implies &\ln 2≥1 \\ 
\implies &e≤2 \\
&\text{A contradiction.}\end{align}$$
A: 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 $x \geq 0$.
A: 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.
A: 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 $k<e$, 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 $b=0.770112$.
