Calculus Volume 1

Chapter 6 | Applications of Integration

725

iv. Note that () d dx ln( Furthermore, () d dx (

r −1 x r =

x r ) = rx

r x .

r ln x ) = r x .

Since the derivatives of these two functions are the same, by the Fundamental Theorem of Calculus, they must differ by a constant. So we have () ln( x r ) = r ln x + C

for some constant C . Taking x =1, we get

() ln(1 r ) = r ln(1)+ C 0 = r (0)+ C C = 0.

Thus ln( x r ) = r ln x and the proof is complete. Note that we can extend this property to irrational values of r later in this section. Part iii. follows from parts ii. and iv. and the proof is left to you. □ Example 6.37 Using Properties of Logarithms

Use properties of logarithms to simplify the following expression into a single logarithm: ln9−2ln3+ln ⎛ ⎝ 1 3 ⎞ ⎠ . Solution We have ln9−2ln3+ln ⎛ ⎝ 1 3 ⎞ ⎠ = ln ⎛ ⎝ 3 2 ⎞ ⎠ −2 ln3+ln ⎛ ⎝ 3 −1 ⎞

⎠ =2ln3−2ln3−ln3=−ln3.

6.37

Use properties of logarithms to simplify the following expression into a single logarithm: ln8−ln2−ln ⎛ ⎝ 1 4 ⎞ ⎠ .

Defining the Number e Now that we have the natural logarithm defined, we can use that function to define the number e .

Definition The number e is defined to be the real number such that

ln e =1.

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