Inverse Hyperbolic Cosine in Terms of ln
Date: 05/27/2003 at 07:37:17 From: Anonymous Subject: Logarithm equations without a rule to govern them! What is t in the equation ln(e^0.1t + 9e^-0.1t) = ln10? There are no rules governing an "expansion" of the ln, or even how to do ln9e^-0.1t I know from just subsituitng some numbers that ln(e^0.1t + 9e^-0.1t) does not = lne^0.1t + lne^-0.1t.
Date: 05/27/2003 at 08:26:08 From: Doctor Mitteldorf Subject: Re: Logarithm equations without a rule to govern them! Hi there, I understand your frustration, but I suggest that it derives from a simple truth that none of your instructors has yet revealed to you: The vast majority of equations that you might write down do not admit of an algebraic solution. This fact is occluded in the educational process by the (understandable) tendency of teachers to focus on those problems that DO have algebraic solutions. This leaves most students with the impression that ALL algebraic equations have algebraic solutions, and consequently that a failure to discover a solution for a given equation reflects poorly on their own ability as mathematicians. In fact, the equation that you have written > ln(e^0.1t + 9e^-0.1t) = ln10 cannot be solved for t. You can raise e to the power of each side e^0.1t + 9e^-0.1t = 10 But this is as far as you can go. One way is to guess values of t, evaluate the LHS, and see how close you get to 10, then use the disparity to help with your next guess. This is called "Newton's Method" or the Newton-Raphson method. Inventing an Operation to Solve x^x = y http://mathforum.org/library/drmath/view/54586.html If you didn't have the 9, however, there would be a combination of symbols that is conventionally notated as the "hyperbolic cosine" or cosh, defined as cosh(x) = (e^x + e^-x)/2 If you take your equation (without the 9) and raise e to the power of each side, you have cosh(0.1t) = 20 This gives a solution of the equation as an inverse hyperbolic cosine. In one sense, this is an answer; but in another sense it just assigns a name to an iterative calculation that arises because there is no algebraic solution. The other equation you offer can be solved algebraically: ln9e^-0.1t = ln10 9 * e^-0.1t = 10 e^-0.1t = 10/9 e^0.1t = 9/10 0.1*t = ln(9/10) t = 10 * ln(9/10) - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/
Date: 05/27/2003 at 09:50:38 From: Doctor Peterson Subject: Re: Logarithm equations without a rule to govern them! Hi. I'd like to add a bit to Dr. Mitteldorf's excellent answer. He transformed your equation to the form e^0.1t + 9e^-0.1t = 10 and mentioned that a special function has been defined, cosh(x) = (e^x + e^-x)/2 I notice that your equation (with the 9) can be rewritten using the hyperbolic cosine: 1/3 e^0.1t + 3 e^-0.1t = 10/3 [dividing by 3) e^(0.1t - ln(3)) + e^(-0.1t + ln(3)) = 10/3 [moving coefficients] [e^(0.1t - ln(3)) + e^(-0.1t + ln(3))]/2 = 5/3 [dividing by 2] cosh(0.1t - ln(3)) = 5/3 [using the function] 0.1t - ln(3) = cosh^-1(5/3) [taking the inverse] t = 10[ln(3) + cosh^-1(5/3)] [solving] As he pointed out, cosh is just a new name given to something that otherwise would have to be solved by numerical methods, so in a sense this is not really an algebraic solution; but the fact that we can use algebra to manipulate a new equation into the form of one that has been given a name means that we have actually accomplished something. This is not uncommon in more advanced applications of algebra or calculus: we find a whole class of equations that can be solved using a single new function (or group of related functions, such as the hyperbolic functions), so that by making tables (in the old days) or writing a program to evaluate the newly introduced functions, we can solve many problems. So a lot of the algebra you do in solving problems like this is to transform your problem into the "standard" form for its class (once you have recognized what that is). Other examples of such special functions used in solving classes of problems include the Bessel functions, the Elliptic integrals, and so on. The logarithms and the trigonometric functions are more familiar examples. I'm not sure I would have noticed that cosh could be used to solve this problem without Dr. Mitteldorf pointing it out. (I hope I would, but...) There can be a lot of "art" involved in this sort of thing; it's like being lost at sea and having to recognize by the smell of the air which group of tropical islands is nearest, and finding a way to head in that direction, rather than just following a compass home. When you finally land on Hyperbolia the natives can care for you. It would be nice if there were a simple procedure that would always get you "home" without having to use what amounts to intuition to see what direction to go in; but despite whatever we say to kids to encourage them, it remains true that "math is hard"! It's only the basic problems (which tend to be taught in school) that are "easy." Fortunately, in modern life there are plenty of tools to solve equations for you numerically, so if you understand the basics, these hard parts can be done automatically. We don't all have to be South Seas navigators. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Date: 05/27/2003 at 12:05:35 From: Doctor Peterson Subject: Re: Logarithm equations without a rule to govern them! Hi. Thinking again about this problem, I realized that the inverse hyperbolic cosine CAN be expressed algebraically in terms of ln, so although our general advice about non-algebraic solutions is valid, it is not necessary here. We CAN solve your problem without resorting to a calculator that knows cosh^-1. Here is how to derive an equation for cosh^-1: Suppose we know that cosh(u) = v That is, e^u + e^-u = 2v Then we can multiply through by e^u: e^(2u) + 1 = 2v e^u This can be written as a quadratic equation in e^u: (e^u)^2 - 2v (e^u) + 1 = 0 Now we can solve for e^u using the quadratic formula: e^u = [2v +/- sqrt(4v^2 - 4)]/2 = v +/- sqrt(v^2 - 1) u = ln(v +/- sqrt(v^2 - 1) You can either apply this expression for the inverse hyperbolic cosine to what I wrote, or else apply the same technique to solve your problem directly, by multiplying your equation (in exponential form) by e^(0.1t) and writing it as a quadratic. I'm sorry it took so long for us to recognize this solution! Note that, as we said, solutions to problems like this tend to depend on knowing the right tricks; you can see that it is easy even for us to miss a solution if we get our minds on the wrong track. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Date: 05/28/2003 at 07:11:56 From: Anonymous Subject: Thank you (Logarithm equations without a rule to govern them!) To Drs. Mitteldorf and Peterson, You have brightened my day. This site has to be the best thing for people like me who are not "south sea navigators" (as Dr. Peterson said) but who have to do a bit of mathematics in their respective applications (I do medicine - and I've done some physics and more basic maths in the past, but I've never done the "higher level stuff," only dabbled in from time to time. I'm learning it now as I go. I hope everyone thanks you for all your hard work and polite answers! Cheers, A student of the University of Queensland, Brisbane, Australia
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