View
229
Download
0
Category
Tags:
Preview:
Citation preview
EXAMPLE 1 Rewrite logarithmic equations
Logarithmic Form Exponential Form
23 = 8a. =2
log 8 3
40 = 1b. 4
log 1 = 0
=c. 12
log 12 1
=d. 1/4
log –14
121 = 12
4=–11
4
GUIDED PRACTICE for Example 1
Rewrite the equation in exponential form.
Logarithmic Form Exponential Form
34 = 811. =3
log 81 4
71 = 72. 7
log 7 = 1
=3. 14
log 1 0
=4. 1/2
log –532
140 = 1
32=–51
2
EXAMPLE 5 Use inverse properties
Simplify the expression.
a. 10log4 b. 5
log 25x
SOLUTION
Express 25 as a power with base 5.
a. 10log4 = 4
b. 5
log 25x = (52) x
5log
=5
log 52x
2x=
Power of a power property
blog xb = x
blog bx = x
EXAMPLE 6 Find inverse functions
Find the inverse of the function.
SOLUTION
b.
a. y = 6 x b. y = ln (x + 3)
a.
6log
From the definition of logarithm, the inverse ofy = 6 x is y = x.
Write original function.y = ln (x + 3)Switch x and y.x = ln (y + 3)
Write in exponential form.
Solve for y.
=ex (y + 3)
=ex – 3 y
ANSWER The inverse of y = ln (x + 3) is y = ex – 3.
GUIDED PRACTICE for Examples 5 and 6
Simplify the expression.
SOLUTION
10. 8 8log x
8 8log x = x blog bb = x
11. 7
log 7–3x
SOLUTION
7log 7–3x = –3x
alog ax = x
GUIDED PRACTICE for Examples 5 and 6
Simplify the expression.
SOLUTION
12. 2
log 64x
Express 64 as a power with base 2.2
log 64x = (26) x
2log
=2
log 26x
6x=
Power of a power property
blog bx = x
13. eln20
SOLUTION
eln20 = elog 20e = 20 elog xe = x
GUIDED PRACTICE for Examples 5 and 6
Find the inverse of14. y = 4 x
SOLUTION
From the definition of logarithm, the inverse of
4logy = 6 y = x.is
y = ln (x – 5).Find the inverse of15.
y = ln (x – 5)
SOLUTION
Write original function.
Switch x and y.x = ln (y – 5)Write in exponential form.
Solve for y.
=ex (y – 5)=ex + 5 y
ANSWER The inverse of y = ln (x – 5) is y = ex + 5.
EXAMPLE 7 Graph logarithmic functions
Graph the function.
SOLUTION
a. y =3
log x
Plot several convenient points, such as (1, 0), (3, 1), and (9, 2). The y-axis is a vertical asymptote.
From left to right, draw a curve that starts just to the right of the y-axis and moves up through the plotted points, as shown below.
EXAMPLE 7 Graph logarithmic functions
Graph the function.
SOLUTION
b. y =1/2
log x
Plot several convenient points, such as (1, 0), (2, –1), (4, –2), and (8, –3). The y-axis is a vertical asymptote.
From left to right, draw a curve that starts just to the right of the y-axis and moves down through the plotted points, as shown below.
EXAMPLE 8 Translate a logarithmic graph
SOLUTION
STEP 1
Graph . State the domain and range.y =2
log (x + 3) + 1
STEP 2
Sketch the graph of the parent function y = x, which passes through (1, 0), (2, 1), and (4, 2).
2log
Translate the parent graph left 3 units and up 1 unit. The translated graph passes through (–2, 1), (–1, 2), and (1, 3). The graph’s asymptote is x = –3. The domain is x > –3, and the range is all real numbers.
GUIDED PRACTICE for Examples 7 and 8
Graph the function. State the domain and range.
SOLUTION
16. y =5
log x
If x = 1 y = 0,x = 5 y = 1,x = 10 y = 2
Plot several convenient points, such as (1, 0), (5, 1), and (10, 2). The y-axis is a vertical asymptote.
GUIDED PRACTICE for Examples 7 and 8
From left to right, draw a curve that starts just to the right of the y-axis and moves up through the plotted points.
The domain is x > 0, and the range is all real numbers.
GUIDED PRACTICE for Examples 7 and 8
Graph the function. State the domain and range.
SOLUTION
17. y =1/3
log (x – 3)
domain: x > 3,
range: all real numbers
GUIDED PRACTICE for Examples 7 and 8
Graph the function. State the domain and range.
SOLUTION
18. y =4
log (x + 1) – 2
domain: x > 21,
range: all real numbers
EXAMPLE 2 Evaluate logarithms
4loga. 64
b. 5
log 0.2
Evaluate the logarithm.
blogTo help you find the value of y, ask yourself what
power of b gives you y.
SOLUTION
4 to what power gives 64?a. 4
log43 64, so= 3.=64
5 to what power gives 0.2?b. =5–1 0.2, so –1.0.25
log =
EXAMPLE 2 Evaluate logarithms
Evaluate the logarithm.
blogTo help you find the value of y, ask yourself what
power of b gives you y.
SOLUTION
=–31
5 125, so1/5
log 125 =–3.c. to what power gives 125?15
d. 36 to what power gives 6? 361/2 6, so36
log 6= =12
.
d. 36
log 6
c. 1/5
log 125
EXAMPLE 3 Evaluate common and natural logarithms
Expression Keystrokes Display
a. log 8
b. ln 0.3
Check
8
.3
0.903089987
–1.203972804
100.903 8
0.3e –1.204
EXAMPLE 4 Evaluate a logarithmic model
Tornadoes
The wind speed s (in miles per hour) near the center of a tornado can be modeled by
where d is the distance (in miles) that the tornado travels. In 1925, a tornado traveled 220 miles through three states. Estimate the wind speed near the tornado’s center.
93 log d + 65s =
EXAMPLE 4 Evaluate a logarithmic model
SOLUTION
= 93 log 220 + 65
Write function.
93(2.342) + 65
= 282.806
Substitute 220 for d.
Use a calculator.
Simplify.
The wind speed near the tornado’s center was about 283 miles per hour.
ANSWER
93 log d + 65s =
GUIDED PRACTICE for Examples 2, 3 and 4
Evaluate the logarithm. Use a calculator if necessary.
2 to what power gives 32?
25 32, so= 5.=2
log 32
27 to what power gives 3?13
271/3 3, so27
log 3= = .
2log5. 32
SOLUTION
27log6. 3
SOLUTION
GUIDED PRACTICE for Examples 2, 3 and 4
Evaluate the logarithm. Use a calculator if necessary.
Expression Keystrokes Display
7. log 12
8. ln 0.75
Check
12
.75
1.079
–0.288
101.079 12
0.75e –0.288
GUIDED PRACTICE for Examples 2, 3 and 4
WHAT IF? Use the function in Example 4 to estimate the wind speed near a tornado’s center if its path is 150 miles long.
9.
= 93 log 150 + 65
Write function.
93(2.1760) + 65
= 267
Substitute 150 for d.
Use a calculator.
Simplify.
93 log d + 65s =
SOLUTION
The wind speed near the tornado’s center is about 267 miles per hour.
ANSWER
EXAMPLE 1 Use properties of logarithms
b. 4
log 21
a. 4
log 37 = 3 –
4log 7
4log
= –0.612
=4
log (3 7)
= 34
log + 74
log
= 2.196
0.792 1.404–
0.792 1.404+
34
logUse 0.792 and4
log 7 1.404 to evaluate thelogarithm.
Quotient property
Simplify.
Use the given values of 34
log7.
4log
and
Write 21 as 3 7.
Product property
Use the given values of 34
log7.
4log
and
Simplify.
EXAMPLE 1 Use properties of logarithms
c. 4
log 49
34
logUse 0.792 and4
log 7 1.404 to evaluate thelogarithm.
72=4
log
= 2.808
2(1.404)
Write 49 as 72
Power property
Use the given value of 7.4
log
Simplify.
4log= 2 7
GUIDED PRACTICE for Example 1
2. 6
log 40
= 5 –6
log 86
log
= –0.263
=6
log (8 5)
= 86
log + 56
log
= 2.059
1.161 0.898+
Quotient property
Simplify.
0.898 1.161– Use the given values of 56
log8.
6log
and
Write 40 as 8 5.
Product property
Use the given values of 56
log8.
6log
and
Simplify.
1. 586
log
56
logUse 0.898 and 8 1.161 to evaluate thelogarithm.
6log
GUIDED PRACTICE for Example 1
56
logUse 0.898 and 8 1.161 to evaluate thelogarithm.
6log
6log3. 64 82=
6log
= 2.322
= 2 86
log
2(1.161)
Write 64 as 82
Power property
Use the given value of 8.6
log
Simplify.
4. 6
log 125 53=6
log
= 2.694
= 3 56
log
3(0.898)
Write 125 as 53
Power property
Use the given value of 5.6
log
Simplify.
EXAMPLE 2 Expand a logarithmic expression
Power property
Expand6
log 5x3
y
6log 5x3
yQuotient property
= 56
log x3 y6
log –6
log+
= 56
log x y6
log –6
log+ 3
Product property
SOLUTION
= 5x3 y6
log –6
log
EXAMPLE 3 Standardized Test Practice
Quotient property
Product property
Simplify.
SOLUTION
–log 9 + 3log2 log 3 = –log 9 + log23 log 3 Power property
= 24log
= log (9 )23 – log 3
= log 9 23
3
The correct answer is D.ANSWER
GUIDED PRACTICE for Examples 2 and 3
Expand5. log 3x4 .
SOLUTION
= log 3 + log x4log 3x4
= log 3 + 4 log x Power property
Product property
GUIDED PRACTICE for Examples 2 and 3
SOLUTION
Quotient property
Product property
Simplify.
= –ln 4 + ln 33 ln 12 Power property
= ln (4 )33 – ln 12
= ln 4 33
12
= ln 9
Condense ln 4 + 3 ln 3 – ln 12.6.
ln 4 + 3 ln 3 – ln 12
EXAMPLE 4 Use the change-of-base formula
SOLUTION
3log 8Evaluate using common logarithms and natural
logarithms.
Using common logarithms:
Using natural logarithms:
3log 8 =
log 8log 3
0.90310.4771
1.893
3log 8 =
ln 8ln 3
2.07941.0986
1.893
EXAMPLE 5 Use properties of logarithms in real life
For a sound with intensity I (in watts per square meter), the loudness L(I) of the sound (in decibels) is given by the function
= logL(I) 10 I
0I
Sound Intensity
0Iwhere is the intensity of a barely audible sound
(about watts per square meter). An artist in a recording studio turns up the volume of a track so that the sound’s intensity doubles. By how many decibels does the loudness increase?
10–12
EXAMPLE 5 Use properties of logarithms in real life
Product property
Simplify.
SOLUTION
Let I be the original intensity, so that 2I is the doubled intensity.
Increase in loudness = L(2I) – L(I)
= log10 I
0I
log10 2I
0I
–
I
0I
2I
0I= 10 loglog –
= 210 log log I
0I
–log I
0I
+
ANSWER The loudness increases by about 3 decibels.
10 log 2=
3.01
Write an expression.
Substitute.
Distributive property
Use a calculator.
GUIDED PRACTICE for Examples 4 and 5
Use the change-of-base formula to evaluate the logarithm.
5log 87.
SOLUTION
5log 8 =
log 8log 5
0.90310.6989
1.292
8log 148.
SOLUTION
8log 14 =
log 14log 8
1.1460.9031
1.269
GUIDED PRACTICE for Examples 4 and 5
Use the change-of-base formula to evaluate the logarithm.
26log 99.
SOLUTION
SOLUTION
=log 30log 12
1.47771.076
1.369
0.95421.4149
0.67426
log 9
10. 12
log 30
12log 30
=log 9
log 26
GUIDED PRACTICE for Examples 4 and 5
WHAT IF? In Example 5, suppose the artist turns up the volume so that the sound’s intensity triples. By how many decibels does the loudness increase?
11.
SOLUTION
Let I be the original intensity, so that 3I is the tripled intensity.
= logL(I) 10 I
0I
GUIDED PRACTICE for Examples 4 and 5
Product property
Simplify.
=Increase in loudness L(3I) – L(I)
= log10 I
0I
log10 3I
0I
–
= 310 log log I
0I
–log I
0I
+
ANSWER
The loudness increases by about 4.771 decibels.
10 log 3=
4.771
Write an expression.
Substitute.
Distributive property
Use a calculator.
= 10 log I
0I
log 3I
0I
–
Recommended