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

1 Answer
Dec 2, 2015

Answer:

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]}