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Logarithms' Relation to Exponents

Date: 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 

 (1')  log_a (X*Y) = log_a (X) + log_a (Y) .

Exponent rule (2)  becomes

   X/Y = a^x / a^y = a^(x-y)


   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   
Associated Topics:
High School Logs

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