Question

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

Answer 1

`x = 0.04979, 1.50499`

Explanation:

We have:

`lnx=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 lt alpha lt 1` and `1 lt beta lt 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(x)` and use an iteration `x_(n+1) = g(x_n)`

There will many functions, `g(x)`, that we can choose, some may work## and converge, others may not work or diverge.

Root 1: For `1 lt x lt 2`

We could try:

`lnx = x^3-3 => x^3=3+lnx`
`:. x = root(3)(3+lnx)`

So we will try the iterative equation:

`x_0 \ \ \ \ = 1.5`
`x_(n+1) = root(3)(3+lnx_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 lt x lt 1`

`lnx = x^3-3 => x = e^(x^3-3)`

So we will try the iterative equation:

`x_0 \ \ \ \ = 0.5`
`x_(n+1) = e^((x_n)^3-3)`