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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

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

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

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$$f[x)=\ln(1+x)-2x \implies \frac{1}{1+x}-2=\frac{-1-2x}{1+x}=0, x=-1/2,f''(-1/2)<0$$ 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.

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