How do you solve # Log_10x = x^3 - 3#?
1 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_(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#
#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#
#a_1 = 0.00100000000230258523#
#a_2 = 0.00100000000230258523#
Let's look at the graph of
graph{x^3 -3 - log x [-7.02, 7.023, -3.51, 3.51]}