Date: 08/22/2001 at 09:28:06 From: Cedric Subject: Antilogs If log sub(a)10 = 0.250, then log sub(10)a equals ? Breakdown: log sub(a)10 = 0.250 The solution requires me to the take the base(a) antilogarithm of both sides. That would be 10 = a^0.250. I would like to know why is this the final answer.
Date: 08/22/2001 at 12:29:05 From: Doctor Peterson Subject: Re: Antilogs Hi, Cedric. I'm not sure I understand; what you show is NOT the final answer, because it doesn't say what log_10(a) is. I'm assuming you understand how you got that far, but don't know where to go from here. We started with log_a(10) = 0.25 and raised a to the power on each side: 10 = a^0.25 You can solve for a by raising both sides to the 4th power: 10^4 = a^(0.25 * 4) a = 10^4 Now take the base-10 log of both sides: log_10(a) = log_10(10^4) = 4 This is an interesting result; let's see in general how log_a(b) and log_b(a) are related. I'll start with the base-conversion formula, taking "log" without a base as any convenient log, such as base-10 or natural log: log_a(b) = log(b)/log(a) log_b(a) = log(a)/log(b) It follows automatically that log_a(b) = 1/log_b(a) If we had thought of that to start with, the problem would have been obvious. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Date: 08/22/2001 at 12:53:54 From: Cedric Subject: Re: Antilogs Why and how do you raise a to the power on each side? Why is the antilog 10 = a^0.25? That is the question...
Date: 08/22/2001 at 14:19:39 From: Doctor Peterson Subject: Re: Antilogs Hi again! Okay, I made the wrong assumption: you didn't know how you got from log_a(10) = 0.25 to 10 = a^0.25 This is because the antilog IS the power; the exponential function is the inverse of the logarithm: a^log_a(x) = x and log_a(a^x) = x That is, the functions f(x) = a^x and g(x) = log_a(x) are inverses of one another; f(g(x)) = x and g(f(x)) = x for any x. Therefore, we can undo the base-a logarithm by raising a to that power: log_a(10) = 0.25 a^(log_a(10)) = a^0.25 raising a to each power 10 = a^0.25 using the inverse (with x=10) That's all there is to it. For more on antilogs, go to our search page and enter the keyword antilog . Here is one of the answers you'll find: Anti-Logarithms http://mathforum.org/dr.math/problems/elisabeth.02.27.01.html - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum