# Number of solutions of a non elementary equation

I have this equation: $$ln(x+1)=2x$$ clearly for $$x>-1$$. How can I prove that it has just two solutions? Since we can't solve it in an elementary way, I tried graphically, thus I have plotted $$ln(x+1)$$ and $$2x$$, which is quite easy and I can see that there are the solutions $$x_1=0$$ and $$x_2\in(-1,0)$$. But now, how can I be perfectly sure that there isn't another solution $$x_3>0$$ in a simple way?

I tried to think about $$ln(x+1)=2x$$ as $$g(x):=\frac{ln(x+1)}{2x}=1$$ for $$x>0$$ but when it comes to study the sign of the derivative of $$g$$ I get stuck.

Thanks for the help

• The function $f(x)=2x-\ln(x+1)$ has derivative $2-\frac 1{1+x}$. which is $>0$ for $x>-.5$ – lulu Jan 22 at 11:44
• Thank you! I guess I got lost in a very simple task. – batman Jan 22 at 11:47

Hint: Between any two zeros there is a point where the derivative is $$0$$. But the derivative of $$2x-\ln (1+x)$$ has only one zero.

• Perfect, thank you. I don't know why I didn't think this earlier! – batman Jan 22 at 11:48

Write $$f(x)=\log(x+1)-2x$$ and find the extrema to establish a table of variations.

We have

$$f'(x)=\frac1{x+1}-2$$ so it is an easy matter to see that $$f$$ is increasing in $$\left(-1,-\dfrac12\right]$$ and increasing in $$\left[-\dfrac12,\infty\right)$$. This is enough to prove that there are at most two roots.

As $$f(-1^+)<0$$, $$f\left(-\dfrac12\right)>0$$ and $$f(\infty)<0$$, and the function is continuous, we can affirm that there is a root in both intervals. (By inspection $$x=0$$ is one of them.)

$$f[x)=\ln(1+x)-2x \implies \frac{1}{1+x}-2=\frac{-1-2x}{1+x}=0, x=-1/2,f''(-1/2)<0$$ As $$f(-1+h)<0, f(\infty)=-\infty$$ (same sign), there are as even number $$0,2,4,...$$ of real roots of $$f(x)=0$$ in $$(-1,\infty)$$. Since $$f(x)$$ has only one max, $$f_{max}=f(-1/2)=-\ln 2+1>0$$ which is positive. Hence, this Eq. $$\ln(1+x)=2x$$ will have only two real roots.