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