# Solve exponential equation $3{e^x}={2e^{3/2}}x$

I want to solve the following exponential equation:

$$\frac{e^x}{x}=\frac{2e^{\frac 3 2}}{3}$$

Now, I think it is easy enough to see in the head that the answer is $x=\frac 3 2$. Though, I wanted to try to solve it in the standard way. I multiply both sides by $x$:

$$e^x=\frac{2xe^{\frac 3 2}}{3}$$

Now I apply the logarithm function to both sides:

$$\ln(e^x)=\ln\left(\frac{2xe^{\frac 3 2}}{3}\right)$$

Which should be equal to

$$x=\ln\left({2xe^{3/2}}\right)-\ln{3}\rightarrow \dots$$

I can't really seem to get to the solution. Any hints on what I'm doing wrong?

• You can consider the monotonicity of the function $\frac{e^x}{x}$. – Riemann Aug 26 '18 at 13:26
• There is not a general and explicit method to solve this kind of equation $e^x=ax$. – W. mu Aug 26 '18 at 13:27

Transcendental equations usually have no closed-form solution, except in coincidental cases (like yours), and there is no systematic way to solve. So you are doing nothing wrong.

When you can transform the equation in a polynomial one, then you can see if the equation is solvable analytically, as the case of polynomials has been settled. Otherwise, you are on your own.

Anyway, for your particular equation, there is a special function (Lambert's $W$) that you can use: it solves $xe^x=y$ for $x$ and its properties have been studied. Your equation can be turned to that form. (At the same time this "proves" that your equation cannot be solved by other means.)

Also note that numerical solvers can find approximations of the roots and may allow you to recognize the exact value if there is one. But I wouldn't recommend this as a routine approach.

Manipulating the expression to $$-xe^{-x}=-\frac32e^{-3/2}$$ we see that the other solution uses the Lambert $W$ function, and is given by $$x=-W(-\frac32e^{-3/2})\approx 0.626$$

Note that the function $$f(x)=xe^{-x}$$ is strictly decreasing on $[1,\infty)$

Thus it is one-t -one on $[1,\infty)$

$$f(x)=f(3/2) \implies x=3/2$$

That is $$(3/2) e^{-3/2} = xe^{-3/2} \implies x=3/2$$

If you do not want to use Lambert function, consider, after taking logarithms as you did, that you look for the other zero of function $$f(x)=x-\log (x)-\frac{3}{2}+\log (3)-\log (2)$$ $$f'(x)=1-\frac{1}{x}\qquad \text{and}\qquad f''(x)=\frac{1}{x^2} > 0 \qquad \forall x$$ The first derivative cancels at $x=1$ and we have $$f(1)=\log (3)-\frac{1}{2}-\log (2) <0$$ which means that you have two roots : $0 < x_1<1$ and $x_2>1$. You already found the obvious root $x_2=\frac 32$

To approximate the root, make a Taylor expansion around $x=1$ to get $$f(x)=\left(\log (3)-\frac{1}{2}-\log (2)\right)+\frac{1}{2} (x-1)^2+O\left((x-1)^3\right)$$ Ignoring the higher order terms, the smallest root is given by $$x_*=1-\sqrt{1+2 \log (2)-2 \log (3)}$$ Now, start Newton iterations and get the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.5651784460 \\ 1 & 0.6188044605 \\ 2 & 0.6256800893 \\ 3 & 0.6257825118 \\ 4 & 0.6257825342 \end{array} \right)$$ which is the solution for ten significant figures.

We could even have a better approximation using a $[2,2]$ Padé approximant of the function built at $x=1$. Solving the quadratic in numerator would lead to the approximation $$\frac{25+14 \log (2)-14 \log (3)-3 \sqrt{6 \left(7+4 \log ^2(2)+4 \log ^2(3)+16 \log (2)-16 \log (3)-8 \log (2) \log (3)\right)}}{19+2 \log (2)-2 \log (3)}$$ which is $\approx 0.625339$ which is not too bad (I hope and wish).