5 Steps to a 5 AP Calculus AB 2019 - William Ma

MA 3972-MA-Book May 8, 2018 13:46

Review of Precalculus 69

Change of Base Formula

logax=

lnx

lna

wherea>0 anda = 1

Example 1

Sketch the graph off(x)=ln(x−2). Note that the domain off(x)is{x|x> 2 }and that

f(3)=ln(1)=0; thus, thex-intercept is 3. (See Figure 5.3-27.)

y

0 (3, 0) x

y = ln(x − 2)

x = 2

Figure 5.3-27

Example 2

Evaluate:

(a) log 28

(b) log 5

1

25

(c) lne^5

Solutions

(a) Letn=log 2 8 and thus 2n= 8 = 23. Therefore,n=3.

(b) Letn=log 5

1

25

, and thus 5n=

1

25

= 5 −^2. Therefore,n=−2.

(c) You know thaty=exandy=lnxare inverse functions. Thus, lne^5 =5.

Example 3

Express ln(x(2x+5)^3 ) as the sum and multiplication of logarithms.

ln(x(2x+5)^3 )=lnx+ln(2x+5)^3 =lnx+3ln(2x+5)

Example 4

Solve 2ex+^1 =18 to the nearest thousandth.

2 ex+^1 = 18

ex+^1 = 9

ln(ex+^1 )=ln 9

x+ 1 =ln 9

x = 1. 197