2.5.1MATHPOWERTM 12, WESTERN EDITION

2.5Chapter 2 Exponents and Logarithms

A logarithmic function is the inverse of an exponential function.

For the function y = 2x, the inverse is x = 2y.

In order to solve this inverse equation for y, we write it in logarithmic form.

x = 2y is written as y = log2x and is read as “y = the logarithm of x to base 2”.

x -3 -2 -1 0 1 2 3 4

y 1

8

1

4

1

21 2 4 8 16

x

y -3 -2 -1 0 1 2 3 4

1

8

1

4

1

21 2 4 8 16

y = 2x

y = log2x

(x = 2y)

2.5.2

y = 2x

y = x

y = log2x

2.5.3

Graphing the Logarithmic Function

The y-intercept is 1.

There is no x-intercept.

The domain is {x | x R}.

The range is {y | y > 0}.

There is a horizontal asymptoteat y = 0.

There is no y-intercept.

The x-intercept is 1.

The domain is {x | x }.

The range is {y | y R}.

There is a vertical asymptoteat x = 0.

y = 2x y = log2x

The graph of y = 2x has been reflected in the line of y = x, to give the graph of y = log2x.

2.5.4

Comparing Exponential and Logarithmic Function Graphs

Logarithms

Consider 72 = 49.

2 is the exponent of the power, to which 7 is raised, to equal 49.

The logarithm of 49 to the base 7 is equal to 2 (log749 = 2).

72 = 49 log749 = 2

Exponential notation

Logarithmic form

In general: If bx = N, then logbN = x.

State in logarithmic form:

a) 63 = 216

b) 42 = 16

log6216 = 3

log416 = 2

State in exponential form:

a) log5125 = 3

b) log2128= 7

53 = 125

27 = 1282.5.5

Logarithms

2.5.6

State in logarithmic form:

y 1

2

x

1.4

log0.5 y x

1.4

1.4log0.5 y x

a) b) 23x2 32

log2 32 = 3x + 2

Evaluating Logarithms

1. log2128

log2128 = x 2x = 128 2x = 27

x = 7

2. log327

log327 = x 3x = 27 3x = 33

x = 3

Note:log2128 = log227

= 7 log327 = log333

= 3

3. log556 = 6 logaam = m

4. log816

log816 = x 8x = 16 23x = 24

3x = 4

5. log81

log81 = x 8x = 1 8x = 80

x = 0

loga1 = 0

2.5.7

x 4

3

6. log4(log338)

log48 = x 4x = 8 22x = 23

2x = 3

7. log 4 83

log 4 83 = x

4x 83

2 2x 23

3

2x = 1

8. 2 log2 8

2 log2 23

= 23

= 8

9. Given log165 = x, and log84 = y, express log220 in terms of x and y.log165 = x

16x = 5 24x = 5

log84 = y8y = 423y = 4

log220 = log2(4 x 5) = log2(23y x 24x) = log2(23y + 4x) = 3y + 4x 2.5.8

Evaluating Logarithms

x 3

2x

1

2

Base 10 logarithms are called common logs.

Using your calculator, evaluate to 3 decimal places:

a) log1025 b) log100.32 c) log102

1.398 -0.495 0.301

Evaluate log29:

log29 = x 2x = 9

log 2x = log 9 xlog 2 = log 9

x log9

log2

x = 3.170

Change of base formula:

loga b log b

log a

2.5.9

Evaluating Base 10 Logs

2.5.10

Evaluating Logs

Given log3a = 1.43 and log4b = 1.86, determine logba.

log3a = 1.43 a = 31.43 log a = 1.43log 3

log4b = 1.86 b = 41.86 log b = 1.86 log 4

logb a log a

log b

logb a 1.43log 3

1.86log 4

logba = 0.609

Suggested Questions:Pages 98-1001-31 odd,33-42, 47,50 a, 52 a

2.5.11