Solving $\ln$ equations with polynomial terms.

Question

How do I solve the following equation for $x$:

$\ln(x) -3x+3=0$

Answer 1

Generally there is not an explicit solution to these equations, meaning a closed form solution in terms of elementary functions. For this reason and to calculate the values of the solutions of such equations, numerical analysis and numerical methods come into play.

But in your specific case, observe that $x=1$ is a solution to the equation. Now, consider the function: $$f(x) = \ln x - 3x + 3.$$ By calculating its derivative and studying the properties of the function (monotonicity), you can conclude whether that particular solution is unique or not.

In a wider aspect, solutions of such equations can be expressed via the Lambert $W$ function. You can find more information here.

Answer 2

$$\ln x-3x+3=0\iff\begin{cases} x>0\\ \ln (xe^{-3x})=-3\end{cases}\iff \begin{cases} x>0\\ xe^{-3x}=e^{-3}\end{cases}\\\iff \begin{cases} x>0\\ -3xe^{-3x}=-3e^{-3}\end{cases}\iff\begin{cases} x>0\\ x=-\frac13W(-3e^{-3})\end{cases}\lor\begin{cases} x>0\\ x=-\frac13W_{-1}(-3e^{-3})\end{cases}$$

$W$ and $W_{-1}$ may be considered because $-3e^{-3}\ge- e^{-1}$. Now, since $-3\le-1$, we have that $W_{-1}(-3e^{-3})=-3$. For the other case, $W(\alpha)$ sends negative numbers to negative numbers and positive numbers to positive numbers, therefore $-\frac13W(-3e^{-3})>0$. Thus

$$x=-\frac{W(-3e^{-3})}{3}\lor x=1$$