logarithm (idea)

Proofs to Jakohn's first three properties:

1. ln(pq) = ln(p) + ln(q)
2. ln(p /q) = ln(p) - ln(q)
3. ln(p ^r) = r ln(p) for every rational number r

using log_e or ln (natural log):

1. If p > 0, then 

(d/dx) (ln px) = (1/px)p = (1/x).

Therefore, ln(px) and ln(x) are both antiderivatives of 1/x (∫(1/x)dx = ln |x| but since x > 0, 
∫(1/x)dx = ln (x) ) so

ln(px) = ln(x) + C

for some constant C. Letting x = 1, we obtain

ln(p) = ln(1) + C.

Since ln(1) = 0, C = ln(p), and therefore

ln(px) = ln(x) + ln(p).

Substituting q for x,

ln(pq) = ln(q) + ln(p).


Example:

ln[(x + 2)(3x - 5)] =

ln(x + 2) + ln(3x - 5)


2. Using the formula ln(pq) = ln(p) + ln(q) with p = 1/q,

ln(1/q) + ln(q) = ln((1/q)*q) = ln(1) = 0

so

ln(1/q) = -ln(q).

Consequently,

ln(p /q) = ln(p*(1/q)) = ln(p) + ln(1/q) = ln(p) - ln(q).


Example:

    x + 2
ln ------- = 
   3x - 5

ln(x + 2) - ln(3x - 5)


3. If r is a rational number and x > 0, then

(d/dx)(ln(xr)) = (1/xr)(d/dx)(xr) = (1/xr)rx(r - 1) = r(1/x) = r/x.

Since ln(xr) and rln(x) are both antiderivatives of r/x,

ln(xr) = rln (x) + C

for some constant C. If we let x = 1

ln(1) = rln(1) + C.

Since ln(1) = 0, C = 0 and, therefore,

ln(xr) = rln(x).


Example:

ln √(x + 1) = 

(1/2)ln(x + 1)

These properties are convenient to use when the derivative is going to be taken. It makes it so that the chain rule, quotient rule, and product rule are unneeded or easier to do.

Consider taking the derivative of the last example without applying the property

(d/dx)(ln(√(x + 1) =

    1      1
-------- * - (x + 1)(-1/2) =
√(x + 1)   2

    1      1      1
-------- * - * ------- =
√(x + 1)   2   √(x + 1)

   1 
--------
2(x + 1)

If the rules are applied to it first, it becomes a simpler problem.

(d/dx)(ln(√(x + 1) =

(d/dx)(1/2)(ln(x + 1) =

1     1        dx
- * ------ * ------ =
2   x + 1      dx

    1
---------
2(x + 1)