2. Logarithmic functions are the inverses of exponential functions,
and any exponential function can be expressed in logarithmic
form.
Similarly, all logarithmic functions can be rewritten in
exponential form.
Logarithms are really useful in permitting us to work with very
large numbers while manipulating numbers of a much more
manageable size.
Every exponential function f(x) = a x, with a > 0 and a ≠ 1. is a
one-to-one function, therefore has an inverse function(f-1).
The inverse function is called the Logarithmic function with base
a and is denoted by loga
Let a be a positive number with a ≠ 1. The logarithmic function with base a,
denoted by loga is defined by:
Loga x = y a y = Х
Clearly, Loga Х is the exponent to which the base a must be raised to give Х
3. y = loga x if and only if x = a y
The logarithmic function to the base a, where a > 0 and a
1 is defined:
2
416
exponential
form
logarithmic
form
Convert to log
form:
216log4
Convert to
exponential form:
3
8
1
log2
8
1
2 3
When you convert an exponential to log form, notice that the
exponent in the exponential becomes what the log is equal to.
4. f(x) = 10x is an exponential function where the base is 10 and
the exponent is x
Let us write this as: y = 10x
Here the power x is the input and the quantity y is the
output.
The domain is the set of x-values and the range is the set of y-
values.
A similar statement made by using the quantity y as the input
and the power x as the out put is called a logarithmic
statement.
When you input a quantity y, what will be the power of the
base 10 to obtain y?
The answer is x
To write this in the proper function form, we exchange x and y.
The statement y = log10x is called a logarithmic function.
Log10y = x
The logarithm of y to the base 10 is x
5. Find the value of: 5log 5
5log 5 x
is obtained by raising the base tThe qua o the pn otit 55y wer x
5 5x
1
2
5 5x
1
2
x
5log 5
1
2
6. Find the value of: 6 6log 3
6
6is obtained by raising the base 6 to theThe quantit powery log 36x
6log 36
6 x
log6x = log636 Since the bases are the same, x = 36
6 3log 6
6 36
2 2 3 3 3
Evaluate:
1
(a)log 8 log 4 ( )log 27 log 3 ( ) log 81
4
b c
2log 8 4
2log 32
5
3
27
log
3
3log 9
2
1
4
3log 81
3log 3
1
7. Obtain ordered pairs and graph f(x) = log10(x)
x 0. 1 0.2 0.4 0.8 1 2 3 4 5
y -1 -0.7 -0.4 -0.1 0 0.3 0.48 0.6 0.7
0.80.60.40.2
0
1.0 1.2 1.4 1.6 1.8
-0.2
-0.4
-0.6
-0.8
0.2
0.4
0.6
0.8
-1.0
2.0 2.2 2.4 2.8 2.8 3.0
(0.1, -1)
(0.2, -0.7)
(0.4, -0.4)
(0.8, -0.1)
(1, 0)
(2, 0.3)
x = 0 is a vertical asymptote for this graph.
9. f(x) = bx
Domain: (-∞, ∞)
Range: (0, ∞)
g(x) = logbx
Doman: (0, ∞)
Range: (-∞, ∞)
f(x) = 2x
g(x) =log2x
y = x
(1,0)
(0,1)
x
y
10. The graph of g(x) = log2(x – h) + k can be obtained by
shifting the graph of f(x) = log2(x) h units horizontally and
k units vertically.
Use the graph of f(x) = log2(x) to obtain the graph of
g(x) = log2(x – 1) + 2
0-1-2-3-4-5 1 2 3 4 5
-1
-2
-3
-4
1
2
3
4
f(x)
g(x)
Here h = 1 and k = 2
The graph of f(x) = log2(x)
shifts 1 unit to the right and
2 units up
x = 1 is a vertical asymptote.
11. Example:
A sum of $500 is invested at an interest rate 9%per year. Find the
time required for the money to double if the interest is compounded
according to the following method.
a) Semiannual b) continuous
Solution:
(a) We use the formula for compound interest with P = $5000, A (t) =
$10,000r = 0.09, n = 2, and solve the resulting exponential
equation for t.
(Divide by 5000)
(Take log of each side)
(bring down the exponent)
(Divide by 2 log 1.045)
t ≈ 7.9 The money will double in 7.9 years. (using a calculator)
10000
2
09.0
15000
2t
2045.1
2t
21.04521log
2t
045.1log2)(logt
2log1.045log2t
12. (b) We use the formula for continuously compounded interest with P =
$5000, A(t) = $10,000, r = 0.09, and solve the resulting exponential
equation for t.
5000e0.09t = 10,000
e 0.091 = 2 (Divide by 5000)
In e 0.091 = In 2 (Take 10 of each side)
0.09t = In 2 (Property of In)
t=(In 2)/(0.09) (Divide by 0.09)
t ≈7.702 (Use a calculator)
The money will double in 7.7 years.
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