Question

How do you solve `Log_10x = x^3 - 3`?

Answer 1

Solve numerically using Newton's method to find:

`x_1 ~~ 1.468510541627833`

`x_2 ~~ 0.00100000000230258523`

Explanation:

Let:

`f(x) = x^3 -3 - log_10 x=x^3-3-(ln x)/(ln 10)`

Then:

`f'(x) = 3x^2-1/(x ln 10)`

Choose a first approximation `a_0` then iterate using the formula:

`a_(i+1) = a_i - f(a_i)/(f'(a_i))`

`= a_i - (a_i^3-3-log_10(a_i))/(3a_i^2-1/(a_i ln 10))`

Choosing `a_0 = 1` and putting this formula into a spreadsheet, I found:

`a_0 = 1`

`a_1 = 1.779512686040294`

`a_2 = 1.521859848585931`

`a_3 = 1.470479042421975`

`a_4 = 1.468513364194865`

`a_5 = 1.468510541633648`

`a_6 = 1.468510541627833`

`a_7 = 1.468510541627833`

Choosing `a_0 = 0.001` with the same formula, I found:

`a_0 = 0.001`

`a_1 = 0.00100000000230258523`

`a_2 = 0.00100000000230258523`

Let's look at the graph of `f(x)`:

graph{x^3 -3 - log x [-7.02, 7.023, -3.51, 3.51]}