How to solve this equation algebraically?

Question

I've come across this interesting equation which I do not know how to solve. The equation is:

$$e^x+\log x =1$$

I used WolframAlpha to solve it and it got but, it didn't provide any solutions. The answer it showed was $x= 0.512222433033230$. How would one solve the equation algebraically?

Answer 1

alex.jordan already commented about one iteration of Newton method starting at $x_0=\frac 12$.

Higher order methods (such as Halley or Householderr) would give better and better results after a single iteration. For illustration purpose, I give you below the value of the first iterate as a function of $n$, the convergence order of the method $$\left( \begin{array}{cc} n & x_1 \\ 2 & 0.512175747765817 \\ 3 & 0.512223514408342 \\ 4 & 0.512222434043136 \\ \end{array} \right)$$

For sure, repeating the process would very quickly converge to the solution $0.512222433033230$.

Just for your curiosity : another solution could be to approximate the function locally using the so called Padé approximants which are ratios of polynomials of different degrees (do not worry : you will learn about them sooner or later). For example, building the simplest at $x=\frac 12$ would give $$e^x+\log(x) \simeq \frac{\sqrt{e}-\log (2)+\frac{8+12 \sqrt{e}+e-4 \log (2)+\sqrt{e} \log (2)}{2 \left(2+\sqrt{e}\right)} \left(x-\frac{1}{2}\right)} {1+ \frac{\left(4-\sqrt{e}\right) }{2 \left(2+\sqrt{e}\right)}\left(x-\frac{1}{2}\right)}$$ So, for $e^x+\log(x)=1$, the above expression would lead to $$x=\frac{12+9 \sqrt{e}-3 e+4 \log (2)+5 \sqrt{e} \log (2)}{2 \left(4+13 \sqrt{e}+e-4 \log (2)+\sqrt{e} \log (2)\right)}\approx 0.512223702539583$$

Answer 2

You can use this one which will give you an approximate value.$e^{x}=1+\frac{x}{1!}+\frac{x^2}{2!}...\infty$,$log(1+x-1)=log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-..$ you can get the value the if $|x|<=1$ . Hope its clear