On-line Math 21

3.2  Exponential and Logarithmic functions

Proposition 2 For all b , c , and d , c and d > 0 ,

1. log_b(cd) = log_b(c)+log_b(d) ,
2. log_b(c^d) = dlog_b(c) , Here d can be negative.
3. 
   

   logb(c) logb(d) = logd(c).

   This last fact means that you can find the logarithm of c to any base, if you know all logarithms of one base.

Proof: The proofs consist of turning around the rules for exponents. I'll show the first one as an example, then you will do the other ones as (written) exercises.

To show the first of these rules, start by labelling all the relevant expressions:

logb(c) : = C logb(d) : = D, and logb(cd) : = E.

Then, unwind what these expressions mean, by the definition of the logarithms:

logb(c) = C Û c = bC logb(d) = D Û d = bD, and logb(cd) = E Û cd = bE.

But if we look harder at that last equation, and substitute for c and d according to the earlier equations,

cd = bE bCbD = bE.

Finally, use the rules for exponents:

bE = bCbD = bC+D

which implies that the two expoenents have to be the same,

E = C+D.

This is, however, the end, since we recall what these letters meant:

logb(cd) = logb(c)+logb(d).

Copyright (c) 2000 by David L. Johnson.