8.4. ln AND logb 165

Proof: Since exp(ln(x)) = x and ln′ exists, it follows

x ln′ (x) = exp(ln(x)) ln′ (x) = exp′ (ln(x)) ln′ (x)= exp(ln(x)) ln′ (x) = x ln′ (x) = 1

and this proves 8.13. Next consider 8.14.

xy = exp(ln(xy)) , exp(ln(x)+ ln(y)) = exp(ln(x))exp(ln(y)) = xy.

Since exp was shown to be 1-1, it follows ln(xy) = ln(x)+ ln(y) . Next exp(0) = 1 andexp(ln(1)) = 1 so ln(1) = 0 again because exp is 1-1. Let f (x) = ln(xm)−m ln(x) . Thenf (1) = ln(1)−m ln(1) = 0. Also, by the chain rule, f ′ (x) = 1

xm mxm−1−m 1x = 0 and so

f (x) equals a constant. The constant can only be 0 because f (1) = 0. This proves the lastformula of 8.15 and completes the proof of the theorem.

The last formula tells how to define xα for any x > 0 and α ∈ R. I want to stress thatthis is something new. Students are often deceived into thinking they know what xα meansfor α a real number because they have a meaning for α an integer and with a little stretchfor α a rational number. Such deception should never be tolerated in mathematics.

Definition 8.4.3 Define xα for x > 0 and α ∈ R by ln(xα) = α ln(x) . In otherwords, xα ≡ exp(α ln(x)) .

From Theorem 8.4.2 this new definition does not contradict the usual definition in thecase where α is an integer.

From this definition, the following properties are obtained.

Proposition 8.4.4 For x > 0 let f (x) = xα where α ∈ R. Then f ′ (x) = αxα−1.Also xα+β = xα xβ and (xα)β = xαβ .

Proof: First consider the claim about the sum of the exponents.

xα+β ≡ exp((α +β ) ln(x)) = exp(α ln(x)+β ln(x))

= exp(α ln(x))exp(β ln(x))≡ xα xβ .

ln((xα)β

)= β ln(xα) = αβ ln(x) , ln

(xαβ

)= αβ ln(x) .

The claim about the derivative follows from the chain rule. f (x) = exp(α ln(x)) and so

f ′ (x) = exp(α ln(x))α

x≡ α

xxα = α

(x−1)xα = αxα−1.

Definition 8.4.5 Define logb for any b > 0,b ̸= 1 by logb (x)≡ln(x)ln(b) .

Proposition 8.4.6 The following hold for logb (x) .

1. blogb(x) = x, logb (bx) = x.

2. logb (xy) = logb (x)+ logb (y)

3. logb (xα) = α logb (x)