Number of solutions to an equation $6\ln(x^2+1)-e^x = 0$

Question

I need to determine how many solutions the equation $6\ln(x^2+1)-e^x = 0$ has. I wanted to find monotonic intervals of this function and check function's values at local extrema so I calculate determinant of the function, which is $\dfrac{12x}{x^2+1} - e^x$, but I have no idea how to solve any of these equations. Can you help me?

Answer 1

For $x\le0,~$ $~6\ln(x^2+1)$ is strictly decreasing, and $e^x$ is strictly increasing, so, if they ever meet, they can only meet once. And meet they must, since the former decreases from $\infty$ to $0$, while the latter increases from $0$ to $1$, so their two paths inadvertently cross.

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For $x\in(0,1)$, approximate both functions with their Taylor series, and solve the corresponding quadratic equation. Use $\ln(1+t)\simeq t$ and $e^u\simeq1+u$.

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For $x\ge1$, one function is strictly convex, while the other is strictly concave, so they can meet no more than twice. But in order for this to happen, the convex function must start at a value superior to that of the concave one, which isn't the case here, since $6\ln2>e^1$. So they can only meet at most once.

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Thus we have shown that there are no more than three roots at the most. Now remember Yves Daoust's comment, which says that there are at least three roots, and draw the conclusion.