Logarithms' Relation to ExponentsDate: 09/29/2001 at 01:35:54 From: Vid Welsh Subject: Proof of logs I have a bunch of rules for logs, properties and suchlike, but I find it hard to remember them without a proof. My precalculus book has no proof of why logs work or even what they are, nor does my calculus book. I understand what logs are, and their relation to Euler's constant, but I don't understand why they are what they are. Please help me. Date: 09/29/2001 at 13:04:28 From: Doctor Fenton Subject: Re: Proof of logs Hi Vid, Thanks for writing to Dr. Math. The way I like to think about logarithms is that they are just another language for describing exponentials. In exponential language, we emphasize the base; logarithms emphasize the exponent. Each property of logarithms is just a property of exponentials, expressed from a new point of view in the logarithmic language. I assume that you know the fundamental exponential properties: if a > 0, then (1) a^m*a^n = a^(m+n) (2) a^m/a^n = a^(m-n) and (3) (a^m)^n = a^(m*n) . If a^x = X (note the capital and lower case letters: x is the exponent, and X is a raised to that exponent), then we can rephrase this as log_a (X) = x , where log_a denotes the logarithm to the base a. This logarithmic statement is exactly a restatement of the exponential relation a^x = X . Logarithmic properties are likewise just restatements of the exponential properties above. To translate, suppose (*) a^x = X and a^y = Y . Restating these relations as logarithms gives (**) log_a (X) = x and log_a (Y) = y. Then exponent rule (1), which says that a^x * a^y = a^(x+y) can be written as X * Y = a^x * a^y = a^(x+y) , which can be restated as log_a (X*Y) = x + y . Substituting for x and y using (**) gives the first rule for logarithms, (1') log_a (X*Y) = log_a (X) + log_a (Y) . Exponent rule (2) becomes X/Y = a^x / a^y = a^(x-y) so log_a (X/Y) = x - y , or (2') log_a (X/Y) = log_a (X) - log_a (Y) . Finally, exponent rule (3) becomes X^n = (a^x)^n = a^(x*n) , which can be restated as log_a (X^n) = x*n , or (3') log_a (X^n) = n * log_a (X) . Those are the basic properties of logarithms. Any additional properties you need can be derived from these. Does this answer you question, or do you have further questions about logarithms? If so, please write again. - Doctor Fenton, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/