SOLUTION: a^x = 1 + bx a and b are known; find x x=0 is a trivial solution that does not depend on a or b, so this doesn't help. I already have a logarithmic expansion to a solvable

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: a^x = 1 + bx a and b are known; find x x=0 is a trivial solution that does not depend on a or b, so this doesn't help. I already have a logarithmic expansion to a solvable      Log On


   

Question 417286: a^x = 1 + bx
a and b are known; find x
x=0 is a trivial solution that does not depend on a or b, so this doesn't help.
I already have a logarithmic expansion to a solvable cubic equation; and I have a heuristic. If there is nothing more elegant, your conclusion to that effect would be informative.
Thanks.
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
You're given an exponential function (which is transcendental) and an algebraic function in the same equation. Transcendental functions do not combine well with algebraic functions. The best way (other than using a calculator) is to use Newton's method.

Newton's method defines a recursive sequence x%5Bi%5D with an initial "guess" x%5B0%5D. Newton's method says that, to find the zeros of a function f(x) = 0 with a given x%5B0%5D,

x%5Bi%2B1%5D+=+x%5Bi%5D+-+%28f%28x%5Bi%5D%29%2Ff%5E%281%29+%28x%5Bi%5D%29%29

Here, f%5E%281%29+%28x%5Bi%5D%29 represents f'(x[i]), the derivative of the function f%28x%29+=+a%5Ex+-+bx+-+1. Here, the derivative of f(x) is f%5E%281%29%28x%29+=%28ln%28a%29%29a%5Ex+-+b.