Solving $\ln(x+1) = \frac{x}{\sqrt{x^2+1}}$, $x>0$.

Question

I would like to solve for positive $a$ in the following:

$$\ln(a+1) = \frac{a}{\sqrt{a^2+1}}$$

Which would be the $x$-value at which $f(x)=\ln(x+1)$ overtakes $g(x)=x/\sqrt{x^2+1}$. I have made no progress on this problem and a search for the approximation Wolfram|Alpha produces on an inverse symbolic calculator produces no results.

My motivation for this problem is that it seems quite simply stated in that neither function seems strikingly complicated, yet its difficult belies its appearance, and that is interesting.

My question: this is likely very difficult, so I do not ask how to solve this, but rather by what means might such a problem be solved? If someone were to take an educated guess toward finding a closed form for $a$, where might they start? Simple algebraic manipulation is obviously no use, so might there be any useful equivalences to invoke?

Answer 1

This equation being transcendental, you need a numerical method.

However, graphing or using inspection you could notice that the solution is very close to $x=1$. So, perform a series expansion for $$y=\log(x+1)-\frac{x}{\sqrt{x^2+1}}$$ to get $$y=\left(\log (2)-\frac{1}{\sqrt{2}}\right)+\left(\frac{1}{2}-\frac{1}{2 \sqrt{2}}\right) (x-1)+\left(\frac{3}{8 \sqrt{2}}-\frac{1}{8}\right) (x-1)^2+O\left((x-1)^3\right)$$ This is just a quadratic in $(x-1)$ and the solution would be $$x=\frac{5 \sqrt{2}-6+2 \sqrt{18-8 \sqrt{2}+(8 -12 \sqrt{2}) \log (2)}}{3 \sqrt{2}-2}\approx 1.08792$$ while the exact solution, obtained using Newton method, is $1.08828$.

Answer 2

We can see that:

$$t=Sin (tan^{-1} x)=\frac{x}{\sqrt{1+x^2}}$$

$-1≤ t ≤ 1$

Since x>0, then we must consider 0≤t≤ 1.Equating with $Ln(1+x)$ we get:

$$ 0≤Ln(1+x)≤ 1= Ln e$$

$$1≤1+x≤ e$$

$$x≤ e-1$$