1. Exponents and Logarithms

2. Logarithmic Functions
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”.
y = 2x 1 2 4 8 16 1 2 4 8 16
y = log2x
(x = 2y)

3. Graphing the Logarithmic Function
y = x
y = 2x
y = log2x

4. Comparing Exponential and Logarithmic Function Graphs
y = 2x
The y-intercept is 1.
y = x
There is no x-intercept.
The domain is {x | x Î R}.
The range is {y | y > 0}.
y = log2x
There is a horizontal asymptote
at y = 0.
There is no y-intercept.
The x-intercept is 1.
The graph of y = 2x has been reflected in the line y = x,
to give the graph y = log2x.
The domain is {x | x > 0}.
The range is {y | y Î R}.
There is a vertical asymptote
at x = 0.

5. Logarithm/Exponent Forms
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).
Exponential
notation
Logarithmic
form
log749 = 2
72 = 49
In general: If bx = N,
then logbN = x.
State in logarithmic form:
State in exponential form:
a) 63 = 216
log6216 = 3
a) log5125 = 3
53 = 125
b) 42 = 16
log416 = 2
b) log2128= 7
27 = 128

6. Exponent to Logarithmic form
State in logarithmic form: a) b)
log2 32 = 3x + 2

7. Evaluating Logarithms
Note:
log2128 = log227
= 7
log327 = log333
= 3
log2128 = x
2x = 128
2x = 27
x = 7
log327 = x
3x = 27
3x = 33
x = 3
3. log556
= 6
logaam = m
4. log816
5. log81
log816 = x
8x = 16
23x = 24
3x = 4
log81 = x
8x = 1
8x = 80
x = 0
loga1 = 0

8. Evaluating Logarithms6.
log4(log338) 7.
= x
log48 = x
4x = 8
22x = 23
2x = 3
2x = 1
9. Given log165 = x, and log84 = y,
express log220 in terms of x and y. 8.
log165 = x
log84 = y
= 23
= 8
16x = 5
24x = 5
8y = 4
23y = 4
log220 = log2(4 x 5)
= log2(23y x 24x)
= log2(23y + 4x)
= 3y + 4x

9. Evaluating Base 10 Logs
Base 10 logarithms are called common logs.
Using your calculator, evaluate to 3 decimal places:
a) log b) log c) log102
1.398
-0.495
0.301
Evaluate log29:
log29 = x
2x = 9
Change of base formula:
log 2x = log 9
xlog 2 = log 9
x = 3.170

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
logba = 0.609