Doing some exercises I have encountered an equation I could not solve. I know it has a numerical solution of about 1.835 but I am missing the algebraic methods to get to that solution analyticaly.
$$\ln({x})=\frac{1+x}{1+2x} \ \implies \ x=e^{\frac{1+x}{1+2x}}$$
I would like to know what algebraic methods I am missing and which should I learn next to solve equations like this one.
On
By inspection, you know that the zero of function $$f(x)=\log(x)-\frac{x+1}{2x+1}$$ is rather close to $x=2$.
By Taylor around this point $$f(x)= \left(\log (2)-\frac{3}{5}\right)+\sum_{n=1}^\infty (-1)^n \frac{4^n\, n+2\ 5^{n+1} }{10^{n+1}\,n}\, (x-2)^n$$ Truncating to some order $(p+1)$ and using power series reversion $$x=2+\sum_{n=1}^p a_n \,t^n+O(t^{p+1}) \qquad \text{where} \qquad t=f(x)-\left(\log (2)-\frac{3}{5}\right)$$ the first coefficient being $$\left\{\frac{50}{27},\frac{5875}{6561},\frac{479375}{1594323},\frac {4371875}{57395628},\frac{5668796875}{376572715308},\frac{450991484375}{183014339639688}\right\}$$
Since we want $f(x)=0$, for this level of expansion, converted to decimals, $$x_{(6)}=\color{red}{1.8350370674}51$$
On
$$\log(x)=\frac{x+1}{2x+1}$$ $$\frac{x+1}{2x+1}=t \implies x=\frac{1-t}{2t-1}\implies \log \left(\frac{1-t}{2 t-1}\right)=t$$ Exponentiate and reverse $$e^{-t}=\frac{2 t-1}{1-t}$$ The only explicit solution is given in terms of the generalized Lambert function (have a look at equation $(4)$ in the linked paper).
One way is by applying Lagrange reversion and letting $x=\frac1w$:
$$\ln(x)=\frac{x+1}{2x+1}\iff ew=\sqrt[w+2]e\implies w=\sum_{n=1}^\infty\frac{e^{-n}}{n!}\left.\frac{d^{n-1}}{dw^{n-1}}e^\frac nw\right|_2$$
If we wanted to expand $\frac1x$ as a series only, it would diverge. Maclaurin series uses factorial power $n^{(m)}$.
$$\frac{d^{n-1}}{dw^{n-1}}e^\frac nw =\sum_{m=0}^\infty\frac{n^m (-m)^{(n-1)}}{y^{m+n-1}m!}$$
which is just a confluent hypergeometric $\,_1\text F_1(a;b;x)$ or Laguerre $\operatorname L_n^r(x)$ function. Expanding $\operatorname L_n^1(x)$’s integral representation gives us an integral solution:
$$\bbox[4px,border:4px #90EE90 double]{x=\frac1{\frac1e-\sum_\limits{n=1}^\infty(-2e)^{-n}\,_1\text F_1\left(n;2;\frac n2\right)}=\frac1{\frac1{\sqrt e}-\frac1{4\pi}\int_0^{2\pi} e^{\frac{ e^{it}}2+it-\frac12}\ln\left(\frac12(e^{it}+1)e^{\frac{e^{it}}2-it-\frac12}+1\right)dt}}$$
Typing in
RealDigits[1/(1/E-NSum[(-2E)^(-n) Hypergeometric1F1[n,2,n/2],{n,1,7000}])]into Mathematica matches the many first true digits:while the, rounded, integral is shown here
Another way uses the Ramanujan master theorem, which would give the same series expansion or Mellin inversion which would possibly give another integral representation with a similar integrand.