

Solving equations with fractional exponents Newton's Method Exponents
Name: Unknown
Status: N/A
Age: N/A
Location: N/A
Country: N/A
Date: Around 1993
Question:
I have been asked how to solve X^(.6)X=Y by a teacher at a
nearby school. I could not figure out how to solve for X with a fractional
exponent. Can anyone help?
Replies:
Except for very special exponents and equations, there are no
analytic solutions for this type of equation. You have to solve this kind of
equation numerically. However, by doing some plotting, you can find out lots
of things about the solutions. If you plot x^(0.6)x as a function of x, you
will find a maximum at about x=0.8154 and the value for the maximum is about
0.0694. If we restrict our search to positive x, (we do not want complex
solutions) we find that there is no solution if y>0.0694. There are two
solutions if 0 < y < 0.0694 and one solution if y<0. There is an old formula for
calculating the square root of a number y. It is an iterative formula,
meaning that you plug the value of x into it over and over to get the answer.
So, to get the square root of y, do the following over and over until you get
bored. Guess a value of x to start with. Take half of x and add to it half
of y divided by x. Take the resulting value of x and apply it to the formula
again. Keep repeating this, and in about 5 tries for any x, you will have the
square root to many significant figures. For this formula and for negative y,
try the following iterative formula. The new x = (0.4*x^(0.6)y)/(10.6*x^(
0.4)), keep iterating the x value and it should be very convergent. Your
first guess should be bigger than .8154. This comes from calculus and is
related to a method called Newton's method. I hope this is helpful.
Samuel B. Bowen
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Update: June 2012

