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)/(1-0.6*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 Click here to return to the Mathematics Archives

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