Fractional Bases in Exponential and Logarithmic FunctionsDate: 04/06/2009 at 22:23:08 From: nick Subject: Exponential and Logarithmic functions for fractional Bases. I know how to sketch the curve of y = a^x for 0 < a < 1. What I want to know is, is it possible for this to have fractional bases? If yes, what would this curve look like? y = log(a)x for 0 < a < 1 I think that y = a^x would be a reflection of y = log(a)x, 0 < a < 1, but I do not know how to solve to find intersection points. Date: 04/07/2009 at 12:23:22 From: Doctor Ali Subject: Exponential and Logarithmic functions for fractional Bases. Hi Nick! Thanks for writing to Dr. Math. I don't get something in your question. The graph of y = log(a)x is a straight line with slope of log(a). Do you agree? What does this have to do with the logarithmic and exponential functions? - Doctor Ali, The Math Forum http://mathforum.org/dr.math/ Date: 04/07/2009 at 16:51:38 From: nick Subject: Exponential and Logarithmic functions for fractional Bases. It's not a straight line graph. By y = log(a)x, I mean the logarithm of x to the base a. So, where would y = log(a)x intersect its inverse, y = a^x, if the base, a, is such that 0 < a < 1 for both functions? Date: 04/07/2009 at 23:20:15 From: Doctor Ali Subject: Exponential and Logarithmic functions for fractional Bases. Hi Nick! Thanks for writing back to Dr. Math. Now I understand you! To denote the logarithm of x to the base a, mathematicians tend to use this notation: log_a(x) Anyway, we want to solve this equation: log_a(x) = a^x First, raise the base, a, to the power of each side: a^(log_a(x)) = a^(a^x) This simplifies to x = a^(a^x) The roots of this equation cannot be found using elementary methods. We can say that the equation certainly _has_ a root between zero and one. In order to prove this, just assume the contrary. This proof starts by assuming that x is greater than one. If so, a^x will be less than the base, a, while a is less than one, so, a^x will be less than one, too. Similarly, a value in (0,1) raised to another value in the same range will return a third value in the same range. This cannot be equal to x, which was a number greater than one -- a contradiction! Although the roots of such equations cannot be found using known algebraic methods, we can certainly find them using estimation methods. For example, assume that a = 1/2. In this case, we have the equation, x = (1/2)^((1/2)^x) I used Maple to estimate the root of this equation to 225 decimal places, and got x = 0.64118574450498598448620048211482366656282095 7191101755139698797543487491878799762234053693 4991685885923330759169565860161623870766673276 3179162860451957962889389366924650303360974882 6842219643704483597704216867575963173086480... So, you see that we can get close to the roots as much as we desire, but we don't have a closed form for the roots of such equations. Does that make sense? Please write back if you still have any difficulties. - Doctor Ali, The Math Forum http://mathforum.org/dr.math/ Date: 04/08/2009 at 00:56:54 From: nick Subject: Exponential and Logarithmic functions for fractional Bases. Dr. Ali, Salaam. I like the way you used the contradiction. I assume you chose x > 1, and not x < 0, simply because we know the graph of log_a(x) is bounded by the y-axis, i.e., x = 0. Is there a specific reason why the roots are not easily solved? After all, we have the same unknown on both sides, and since they are inverse functions, they intersect at y = x. So essentially we are solving y = x, as well as y = a^x. In other words, x = a^x. Can this one be solved? If not, would it be okay to choose an arbitrary base between 0 and 1 for a, and use the Newton-Raphson Method? I do not have Maple. Thanks, Dr. Ali. I really appreciate all you have done. Date: 04/08/2009 at 10:36:54 From: nick Subject: Exponential and Logarithmic functions for fractional Bases. On another note, Dr. Ali, consider the function sin^n(x). It is the same as (sinx)^n, correct? Why then do I never see ln^n(x)? I only ever see (lnx)^n instead. Thanks. Date: 04/10/2009 at 09:53:47 From: Doctor Ali Subject: Exponential and Logarithmic functions for fractional Bases. Hi Nick! Thanks for writing to Dr. Math. As I wrote earlier, the roots of this equation cannot be found using _elementary_ methods. In fact, there is a non-elementary function that helps you solve the equation for x. It is called the "LambertW" function. To learn more about it, take a look at Wikipedia http://en.wikipedia.org/wiki/Lambert_W_function MathWorld http://mathworld.wolfram.com/LambertW-Function.html Can you find the root of a^x = x using this new function you've learned? About the second part of your question: I would say that root approximation methods such as Newton-Raphson method can be applied for most such equations. You can get closer and closer to the root of this equation using more and more iterations. To estimate the roots of such equations, Maple, MATLAB, Mathematica, Derive, PERL and other computer software all use the same methods. The only difference is that computers do the same process much, much faster. Please write back if you still have any difficulties. - Doctor Ali, The Math Forum http://mathforum.org/dr.math/ Date: 04/12/2009 at 14:44:55 From: nick Subject: Exponential and Logarithmic functions for fractional Bases. Dear Dr. Ali, The first example on the Wikipedia page has a similar function, but not similar enough for me to model my answer. So despite reading both articles fully, I have NOT come up with the solution to x = a^x. I get this far: x = a^x 1 = x/a^x 1 = xe(-xlna) -lna = -xlnae(-xlna) At that point, I get stuck. Can you correct me? Thanks so much for your help. Date: 04/13/2009 at 23:58:34 From: Doctor Ali Subject: Exponential and Logarithmic functions for fractional Bases. Hi Nick! Thanks for writing to Dr. Math. You've done the steps correctly so far. Picking up where you left off: -Ln(a) = -x Ln(a) exp[-x Ln(a)] Here, you may use a change of variables to make the equation more understandable. Let's assume that w = -x Ln(a) If we write the equation in terms of w, we will have -Ln(a) = w exp(w) This is just like the original form of the definition of the LambertW function. By solving the above equation for w, we have w = LambertW[-Ln(a)] Finally, you just need to write w in terms of x again. That is, -Ln(a)x = LambertW[-Ln(a)] So, we will have, LambertW[-Ln(a)] x = - ------------------ Ln(a) Does that make sense? Please write back if you still have any difficulties. - Doctor Ali, The Math Forum http://mathforum.org/dr.math/ |
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