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Solve $ {3}{x}\mathrm{{=}}{e}^{x} $

Solving graphically with the help of a grapher tool we get $x\approx 1.8$ and approx $4.5$.

Which properties of exponential functions tells us that there will be two solutions?

Are there always two solutions when a (linear function )= an (exponential function)?

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    $\begingroup$ Hint: a straight line intersects a convex (or concave) curve at at most $2$ points. $\endgroup$ – dxiv Mar 23 '17 at 5:53
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Let $f(x) = 3x - e^x$. Then $f(0) = -1$ and $f(1) = 3 - e > 0$, so by the IVT $f$ has a zero in $(0,1)$. But $\lim_{x\to\infty}f(x) = -\infty$, so there must be another zero in $(1,\infty)$. On the other hand, $f(x) < 0$ for all $x < 0$, so there are no zeros in $(-\infty,0)$. Thus $f$ has exactly two zeros.

There will not always be two solutions. Consider $x = e^x$ and $x + 1 = e^x$.

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The slope of the tangent at the graph of $f(x)=e^x$ at $x=\log(3)$ is exactly $3$. Since $f(x)$ is a convex function, any line with slope $3$ that lies above the line $y=3(x-\log 3)+3$ meets the graph of $f(x)$ at exactly two points.

Additionally, since $e^x\approx 3+3(x-\log(3))+\frac{3}{2}(x-\log(3))^2$ in a neighbourhood of $x=\log(3)$, the two solutions of $e^x=3x$ are reasonably close to $\log(3)\pm\sqrt{2(\log(3)-1)}$. If we start Newton's method at such points, we find accurate approximations of the actual roots in very few steps. And by Lagrange inversion theorem, the smallest root has the "closed" form $$\zeta_0 = \sum_{n\geq 1}\frac{n^{n-1}}{n! 3^n} $$ that by chance is pretty close to the golden ratio $\frac{-1+\sqrt{5}}{2}$.

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Ethan Alwaise sshowed that there are two roots.

In fact, you will learn sooner or later that any equation which can write $$A+Bx+C\log(D+Ex)=0$$ has (real and/or complex) solution(s) in terms of Lambert function.

Four your case, the solutions are given by $$x_1=-W\left(-\frac{1}{3}\right)\approx 0.619061$$ $$x_2=-W_{-1}\left(-\frac{1}{3}\right)\approx1.51213$$

The Wikipedia page gives series expansion which would allow you to compute the numerical values.

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    $\begingroup$ (+1) I find it odd you did not mention Newton's method for approximating $W$ and $W_{-1}$ close to the origin! $\endgroup$ – Jack D'Aurizio Mar 23 '17 at 6:18

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