Solving Log EquationDate: 04/20/2008 at 11:43:11 From: Jen Subject: Finding log(log(x)) Is there a rule for finding log(log(x)), or for finding log(x + log(x))? I have gotten to this equation: log(x) + log(x+log(x)) + log(x+log(x) + log[x+log(x)]) = 0 I need to be able to solve for x. Date: 04/21/2008 at 14:38:19 From: Doctor Ali Subject: Re: Finding log(log(x)) Hi Jen, Thanks for writing to Dr. Math. There are no general formulas for making log(log(x)) or log(x+log(x)) simpler. However, your equation is interesting. In your equation, the sum of three logarithms is zero. This means that if all of them are not zero, some are positive and some are negative. Where is log(x) positive and where is it negative? (Just remember that by logs, we mean logarithms in base 10.) For x > 1, log(x) > 0 For x < 1, log(x) < 0 For x = 1, log(x) = 0 First, let's assume that x > 1 and analyze the three parts of your equation: *) 1st LOG: The first thing we can deduce is that: log(x) > 0 **) 2nd LOG: We know that x > 1 and also log(x) > 0. What can we conclude? That x + log(x) > 1 since when you add a positive value to something more than one, the result is still more than one. So, x + log(x) > 1 and log(x + log(x)) > 0 ***) 3rd LOG: Let A be x + log(x). The third log will then be log(A + log(A)) We know that A > 1. So, log(A) > 0 If we add the two inequalities together, we get that: A + log(A) > 1 and we can deduce that the third log in the equation is also positive. In summary, when we assume that x > 1, all of the three logs are positive. We know that the sum of three positive values can never be zero. Thus, there are no roots for this equation if we assume that x > 1. You can try a similar approach for x < 1. After considering x > 1 and x < 1, the only value that remains is x = 1. For x = 1, you can just plug x = 1 in the equation and see if it works. Please write back if you still have any difficulties. - Doctor Ali, The Math Forum http://mathforum.org/dr.math/ Date: 04/21/2008 at 16:34:02 From: Jen Subject: Thank you (Finding log(log(x))) Thank you very much for the help. I actually figured this out this morning, but it was good to see that my logic was not faulty. |
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