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

Dec 2, 2015

Solve numerically using Newton's method to find:

${x}_{1} \approx 1.468510541627833$

${x}_{2} \approx 0.00100000000230258523$

#### Explanation:

Let:

$f \left(x\right) = {x}^{3} - 3 - {\log}_{10} x = {x}^{3} - 3 - \frac{\ln x}{\ln 10}$

Then:

$f ' \left(x\right) = 3 {x}^{2} - \frac{1}{x \ln 10}$

Choose a first approximation ${a}_{0}$ then iterate using the formula:

${a}_{i + 1} = {a}_{i} - f \frac{{a}_{i}}{f ' \left({a}_{i}\right)}$

$= {a}_{i} - \frac{{a}_{i}^{3} - 3 - {\log}_{10} \left({a}_{i}\right)}{3 {a}_{i}^{2} - \frac{1}{{a}_{i} \ln 10}}$

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 $f \left(x\right)$:

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