### Lecture Notes 3.3 Logarithms

```3.3
The logarithm as an inverse function
In this section we concentrate on understanding the logarithm function. If the logarithm is understood
as the inverse of the exponential function, then the variety of properties of logarithms will be seen as
naturally flowing out of our rules for exponents.
3.3.1
The meaning of the logarithm
The logarithmic function g(x) = logb (x) is the inverse of an exponential function f (x) = bx . and so the
meaning of y = logb (x) is by = x. The expression by = x is said to be the “exponential form” for the
logarithm y = logb (x). The positive constant b is called the base (of the logarithm.)
Some worked exercises.
Write each of the following logarithms in exponential form and then use that exponential form to
solve for x.
1. log2 (8) = x
Solution. The exponential form is 2x = 8. Since 23 = 8 the answer is x = 3 .
2. log2 (247 ) = x
Solution. The exponential form is 2x = 247 . So x = 47 .
1
3. log2 ( ) = x
2
Solution. The exponential form is 2x =
1
1
. Since 2−1 = the answer is x = −1 .
2
2
1
4. log2 ( ) = x
8
Solution. The exponential form is 2x =
1
1
. Since 2−3 = the answer is x = −3 .
8
8
√
5. log2 ( 3 2) = x
Solution. The exponential form is 2x = 21/3 . So x = 1/3 .
3.3.2
The graph of a logarithm function
The graph of y = 2x was drawn in an earlier lecture (see figure 1.)
Figure 1. The graph of y = 2x
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The graph of the inverse function y = log2 x is obtained by reflecting the graph of y = 2x across the
line y = x.
Figure 4. The graph of y = log2 x
If we draw them together, we have the picture below, in figure 3:
Figure 5. The graphs of y = 2x and y = log2 x (in blue) with the line of reflection, y = x (in green)
We agreed earlier that the exponential function f (x) = bx has domain (−∞, ∞) and range (0, ∞).
Since g(x) = logb x is the inverse function of f (x) the domain of the log function will be the range
of the exponential function, and vice versa. So the domain of g(x) = logb x is (0, ∞) and the range is
(−∞, ∞).
The most useful base for logarithms is e. We will abbreviate loge (x) by ln(x) and speak of the “natural
logarithm”.
Sometimes, for historical reasons, we may use base 10. It is customary to speak then of the “common
logarithm” and abbreviate log10 (x) by log(x), dropping the subscript. However (warning!), in higher
mathematics and engineering applications, log(x) means base e and is equivalent to ln(x).
In these notes we will use log(x) to mean log10 (x).
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One more abbreviation – often in computer science, because computers store data in binary (in bits
of zeroes and ones), one uses base 2. There is now an attempt to write log2 (x) as lb(x) and speak of the
“binary” logarithm.
In summary, here are our abbreviations:
1. ln x means the logarithm base e,
2. log x means the logarithm base 10 and
3. lb x means the logarithm base 2.
A few more worked exercises.
Write each of the following logarithms in exponential form and then use that exponential form to
solve for x.
1. log(1000) = x
Solution. The exponential form is 10x = 1000. Since 103 = 1000 the answer is x = 3 .
1
)=x
e3
Solution. The exponential form is ex = e−3 so the answer is −3 .
2. ln(
1
3. lb( √ ) = x
2
√
1
1
Solution. The exponential form is 2x = √ . Since 21/2 = 2 then 2−1/2 = √ and so the answer
2
2
is x = −1/2 .
3.3.3
Properties of exponential functions in terms of logarithms
The logarithm function plucks the exponent from an expression. For this reason, the properties of
exponents translate into properties of logarithms.
For example, we know that when we multiply two terms with a common base, we add the exponents:
(bx )(by ) = bx+y
(8)
Suppose we call the first term M := bx and the second term N := by . Then one may ask the question,
“What is the exponent on b in the product M N ?
The answer is “We add the exponents appearing in M and N . In other words (if we learn to translate
“logb ” as “the exponent on b that...”), we can restate this exponent property as “when we multiply
numbers we add their exponents”. This is the product property for logarithms:
logb (M N ) = logb M + logb N
(9)
What happens when we divide two terms with a common base?
bx
= bx−y
by
(10)
When we do division, we subtract exponents. So, in the language of logarithms, we have the quotient
property, “the exponent in a quotient is the difference of the two exponents”:
logb (
M
) = logb M − logb N
N
122
(11)
A third important property of exponents: when we raise a term like bx to a power, we multiply exponents.
(bx )c = bxc
(12)
In our “logarithm language” (thinking of M as bx ) we have the exponent property
logb (M c ) = c logb M
(13)
Each of these three properties is merely a restatement, in the language of logarithms, of a property of
exponents.
3.3.4
Changing the base
Suppose we want to change the base of our logarithm. This often occurs when we want to use a “good”
base like e on a problem which began with a different base.
Maybe we want to work with base c but our problem began with base b:
y = logb x.
Rewrite this in exponential form:
by = x.
and take the log of both sides of the equation. If we want to work in base c then let us apply logc () to
both sides of our equation.
logc (by ) = logc (x).
Now we use the exponet property (equation 13, pulling the exponent y outside the logarithm:
y logc (b) = logc (x).
Solve for y:
y=
logc x
.
logc b
What we have discovered is that y has been rewritten. So y, which was originally equal to logb x is now
logb x =
logc x
logc b
(14)
Let’s call this the “change of base” equation.
Example. Suppose we want to compute log2 (17) but our calculator only allows us to use the natural
logarithm ln. Then, by the change of base equation (equation 14), we can write
log2 (17) =
3.3.5
ln 17
≈ 4.087463.
ln 2
More on the logarithm as an inverse function
We began this lecture by defining g(x) = logb (x) as the inverse function of f (x) = bx . Since these
functions are inverses, we know then that
(f ◦ g)(x) = (g ◦ f )(x) = x.
Let us examine this in more detail.
123
(15)
Note that (g ◦ f )(x) = g(f (x)) = g(bx ) = logb (bx ). So, since the log function and the exponential
function are inverse functions, this must be equal to just x and so we have
logb (bx ) = x.
(16)
This equation is really fairly easy to understand. If we translate “logb (x)” as “the exponent on b that
give x” then we should translate logb (bx ) as “the exponent on b which gives bx .” Obviously this should
be x since x is the exponent one places on b to get bx . (If that doesn’t make sense, read through it one
more time slowly....)
Since (f ◦ g)(x) = x we also have x = (f ◦ g)(x) = f (g(x)) = f (logb x) = blogb x . So
blogb x = x.
(17)
This is almost as easy to understand as equation 16. It says that if we place on b “the exponent you put
on b to get x” (logb x) then we should just get x!
We will explore these properties more in the next section.
Once we understand the logarithm as the inverse of the exponential function, we are prepared to find
the inverse of many functions involving the logarithm. Here are some examples.
Some worked examples on inverse functions Find the inverse function of f (x).
2
1. f (x) = ex .
2. f (x) = ex
2
−5
.
3. f (x) = 5 + ex .
4. f (x) = log2 (x + 2) + 2.
Solutions
2
2
1. To find the inverse of f (x) = ex set y = ex and then swap inputs and outputs so that
2
x = ey .
Take the natural logarithm of both sides
ln x = y 2
and solve for y by taking square roots of both sides
√
ln x = y.
So one inverse function is
f −1 (x) =
2. To find the inverse of y = ex
2
−5
√
ln x .
we swap letters so that x = ey
ln x = y 2 − 5,
add 5 and take square roots so that
f −1 (x) =
124
√
ln x + 5 .
2
=5
, take natural logs of both sides
3. To find the inverse of y = 5 + ex we swap variables, subtract 5 from both sides and then take the
natural log to get ln(x − 5) = y. So
f −1 (x) = ln(x − 5) .
4. To find the inverse of f (x) = log2 (x + 2) + 2 we write
y = log2 (x + 2) + 2,
change variables (to indicate that we are swapping inputs and outputs)
x = log2 (y + 2) + 2,
and subtract 2 from both sides
x − 2 = log2 (y + 2).
At this point it is best to write this logarithmic equation in exponential form.
2x−2 = y + 2.
Subtract 2 from both sides
2x−2 − 2 = y
and then write out our answer using inverse function notation.
f −1 (x) = 2x−2 − 2
3.3.6
Other resources for logarithmic functions
In the free textbook, Precalculus, by Stitz and Zeager (version 3, July 2011, available at stitz-zeager.com)
this material is covered in sections 6.2 and 6.3.
In the free textbook, Precalculus, An Investigation of Functions, by Lippman and Rassmussen (Edition
1.3, available at www.opentextbookstore.com) the logarithms are covered in section 4.3.
In the textbook by Ratti & McWaters, Precalculus, A Unit Circle Approach, 2nd ed., c. 2014 this
material appears in section 3.2. In the textbook by Stewart, Precalculus, Mathematics for Calculus, 6th
ed., c. 2012 (here at Amazon.com) this material appears in in section 4.3.
There are lots of online resources on exponential functions. Here are some I recommend.
1. Dr. Paul’s online math notes on logarithms.
2. Videos on logarithms from Khan Academy,
Homework.
As class homework, please complete Worksheet 3.3, Exponential Functions available through
the class webpage.
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