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 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
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.
 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
 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
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
 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.
f(x) = 2x g(x) = logx2
-2 0.25 0.25 -2
-1 0.5 0.5 -1
0 1 1 0
1 2 2 1
2 4 4 2
3 8 8 3
f(x) = 2x
g(x) =log2x
y = x
(1,0)
(0,1)
x
y
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
 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.
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
(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.
Call us for more
information:
www.iTutor.com
1-855-694-8886
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Logarithm

  • 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.
  • 8. f(x) = 2x g(x) = logx2 -2 0.25 0.25 -2 -1 0.5 0.5 -1 0 1 1 0 1 2 2 1 2 4 4 2 3 8 8 3 f(x) = 2x g(x) =log2x y = x (1,0) (0,1) x y
  • 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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