2^x = x^2Date: 02/13/2002 at 23:26:33 From: Jeremy Kelly Subject: Exponents Recently I was trying to do some random math problems to get a little smarter when I came across something that I couldn't do. 2^x = x^2 I noticed first that there were 2 roots (2 and 4) but there was one that I didn't know of that was negative. Thinking that this was too obvious, I switched the problem to 5^x = x^2 and tried to solve. I took the ln of both sides and tried from there, but I didn't really know where to go. Using some big 'o' and such, I deduced that x had to be negative and there was only one Real value for x that worked, (and graphing proved that), but I didn't know how to find it without graphing it. If you could help that would be great. Jeremy Kelly Date: 02/14/2002 at 09:24:29 From: Doctor Rick Subject: Re: Exponents Hi, Jeremy. I don't know of a way to solve your problem analytically, but I know a method by which we can find successively better approximations to the solution. It's called the Newton-Raphson method, and it can be understood from graphical considerations. Here is an explanation: Cow Grazing Half the Circle: Newton-Raphson Method http://mathforum.org/dr.math/problems/julus1.18.98.html Let's change your problem 2^x = x^2 for what x? into a function: f(x) = 2^x - x^2 Now the question is, for what x is f(x) = 0; or, what are the roots of f(x)? The Newton-Raphson method starts with some first guess, x[0], and finds the next guess, x[1], by a formula. Then, using this guess, we apply the same formula to find a new guess, x[2]. We continue until we're as close as we wish. The formula is x[i+1] = x[i] - f(x[i])/f'(x[i]) We need f'(x), the derivative of f(x). It is f'(x) = 2^x * ln(2) - 2x Thus the formula for our problem is x[i+1] = x[i] - (2^x[i]-x^2)/(2^x[i]*ln(2)-2x) You can set this up in a spreadsheet. Then try different first guesses x[0]. You'll find that the algorithm zeroes in on one of the three roots, depending on the starting value. If I start with x[0] = 0, I get the root: x = -0.766664696 after 5 iterations. You can verify: 2^-0.766664696 = 0.587774756 (-0.766664696)^2 = 0.587774756 If I start with x[0] = 1, I get the root x=2. If I start with x[0] = 3, I get the root x = 4. You have observed that there are three roots. I'll leave the variant with 5^x, and others, up to you. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
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