How do you solve ln(x) = x^3 - 3?

Sep 17, 2017

$x = 0.04979 , 1.50499$

Explanation:

We have:

$\ln x = {x}^{3} - 3$

This equation cannot be solved analytically, so first we graph the functions to get a "feel" for the solutions:

So, we establish that there are two solutions, approximately $0 < \alpha < 1$ and $1 < \beta < 2$, which we attempt to find numerically. We could use Newton-Rhapson but as this question is posed at precalculus level, let us instead use an iterative approach, by rearranging the equation into the form:

$x = g \left(x\right)$ and use an iteration ${x}_{n + 1} = g \left({x}_{n}\right)$

There will many functions, $g \left(x\right)$, that we can choose, some may work and converge, others may not work or diverge.

Root 1: For $1 < x < 2$

We could try:

$\ln x = {x}^{3} - 3 \implies {x}^{3} = 3 + \ln x$
$\therefore x = \sqrt[3]{3 + \ln x}$

So we will try the iterative equation:

${x}_{0} \setminus \setminus \setminus \setminus = 1.5$
${x}_{n + 1} = \sqrt[3]{3 + \ln {x}_{n}}$

Using excel we can quickly process the iterative equation to any degree of accuracy. Here we work to 5dp:

Put $a = 1.5$:

As it happens, and attempt of using ${x}_{0}$ near $0$ also converges on this root, so for the other root we will attempt a different configuration of the iterative function.

Root 2: For $0 < x < 1$

$\ln x = {x}^{3} - 3 \implies x = {e}^{{x}^{3} - 3}$

So we will try the iterative equation:

${x}_{0} \setminus \setminus \setminus \setminus = 0.5$
${x}_{n + 1} = {e}^{{\left({x}_{n}\right)}^{3} - 3}$