How do you solve #2^x = 10x# ?
1 Answer
Mar 4, 2017
There is no pure algebraic solution using elementary functions, but there is an effective numerical method.
Explanation:
Given:
#2^x = 10x#
Define:
#f(x) = 2^x - 10x#
Differentiating, we get:
#f'(x) = 2^x ln 2 - 10#
Using Newton's method, we can choose an initial approximation
#a_(i+1) = a_i - (f(a_i))/(f'(a_i)) = a_i - (2^(a_i)-10a_i)/(2^(a_i)ln 2-10)#
Putting this into a spreadsheet with
Then setting
So there seem to be two solutions:
#x_1 ~~ 0.1077550150002717#
#x_2 ~~ 5.8770105937921375#
Looking at the graphs of
graph{(y-2^x)(y-10x) = 0 [-7, 13, -11.1, 68.9]}