Teaching Logarithms
What is a Logarithm?
Research has found that students lack a foundational understanding of
logarithms. One researcher wrote, "Even after completing a unit on logarithms,
many students continue to struggle to reason about logarithms in powerful
ways, which suggests that they have not developed a useful meaning of
logarithms" (Siebert, 2017). Often, students can solve basic problems that
include logarithms, but when confronted with a statement such as \(log_4 64\)
students do not know how to interpret this statement.
The following section presents a way logarithms can be introduced to
secondary students to better help them conceptualize and understand the
fundamentals.
Consider the following question.
What exponent must we raise 10 by to obtain 54? Or in the language of
mathematics, what value of \(x\) makes \(10^x=54\) true?
We know \(10^1=10\) and \(10^2=100\). What does that tell us about the exponent
required of 10 to obtain 54? Our exponent must be between 1 and 2. How about
1.5? Using a calculator, \(10^{1.5}\) = 31.6228. What does this tell us about
the required exponent? It must be greater than 1.5 but less than 2. Pull out a
calculator
and try some numbers. How many decimal places can you get?
Now, use the calculator to compute \(log54\). How does this number compare
to the one you obtained? Hopefully, they are nearly the same number
(Gamble, 2005).
The reason for this is because we just discovered what a logarithm is. A
logarithm tells us the exponent required of a base to obtain a certain
number. In general,
\(log_a b\) is read as "log base \(a\) of \(b\)" and is the number we must
raise \(a\) by to obtain \(b\).
Let's consider a few examples.
Using the definition provided above, what is \(log_3 27\)? That is the
number we must raise 3 by to obtain 27. What is that
number? \(3^3 = 27\); therefore \(log_3 27 = 3\).
What is \(log_5 \frac{1}{5}\)? Remember that
negative exponents produce fractions. \(5^{-1} = \frac{1}{5}\); therefore, \(log_5 1/5
= -1\).
Here's a third example. What is \(log_9 3\)? In other words, what is the
exponent required of 9 to obtain 3? We know \(\sqrt{9}=3\). Using exponent notation,
\(9^{\frac{1}{2}}=3\). So, what is \(log_9 3\)? It is \(\frac{1}{2}\).
Note that in each of the above examples, the logarithm is equal to a number.
This is always the case. Often, when students are first introduced to
logarithms, they mistakenly believe the logarithm is a separate entity from
the numbers. However, the logarithmic form is simply just another way to
represent a number. Just as 2*2 is another way to write the number 4,
\(log_2 16\) is another way to write the number 4. Remember, a logarithm "is not
three separate entities but rather a single numerical value, which is, in
fact, an exponent" (Wood, 2004). It is important to make this distinction for
students.
It is essential for students to develop this type of reasoning. One
researcher made the bold but accurate claim that "there are very few
logarithmic expressions… that cannot be evaluated by going back to this
basic and fundamental understanding" (Liang & Wood, 2005).
Logarithms as Functions
Research has also found that students struggle to see logarithms as the
output of a function. They do not understand that logarithmic functions by
themselves have a specific domain and range. When asked to draw a
logarithmic function, some students even confuse the logarithmic graph with
the exponential graph. In one study, a student drew both the logarithmic and
exponential graph on the same coordinate axes and stated that both represent
the logarithmic graph (Kenney & Kastberg, 2013). The following section
outlines a way teachers can help students gain an understanding of
logarithms as functions.
Now that we have conceptualized what a logarithm does, let’s look at how
logarithms behave as functions.
A brief review of functions will be important to understanding logarithmic
functions (Kenney & Kastberg, 2013).
Review: A function is a set of ordered pairs \((x,f(x))\). \(x\) is the input and
\(f(x)\), more commonly known as \(y\), is the output. The function describes how \(x\)
relates to \(f(x)\). For example, consider \(f(x)=x+2\). \(x\) is whatever number is put
into the function, and \(f(x)\) is the output. The right side of the equation
tells us what to do to the input to get the output. For this example, the
function tells us to add 2 to the input to get the output. Putting several
values into the function would yield a set of ordered pairs that could then
be graphed to visually represent our function.
Consider the function \(f(x)=log_2 x\). The function tells us to take the
logarithm base 2 of each input \(x\). Using the wording of the previous section,
the function tells us to find what exponent is required of 2 to obtain the
input value. With this understanding, we can create a table of inputs and
outputs for this function.
| \(x\) | \(f(x)\) |
| 1 | \(log_21=0\) |
| \(\frac{1}{2}\) | \(log_2\frac{1}{2}=-1\) |
| 4 | \(log_2 4=2\) |
| 8 | \(log_2 8=3\) |
| -1 | \(log_2 (-1)\) DNE |
| 0 | \(log_2 0\) DNE |
Convince yourself that each of the rows of this table is true for the given
function.
Let's note the last two rows of this table. Why does an input of -1 yield no
output? \(log_2 (-1)\) asks us to find the exponent 2 must be raised by to
obtain -1. What kind of exponent produces negative numbers? None. There is
no number that when acting on a base yields a negative number
(Chua & Wood, 2005). Similarly, there is no exponent that will yield zero.
What do these two facts tell us about the domain? In interval notation, the
domain of this function is \((0,\infty)\).
We can plot these points to investigate the shape and important
characteristics of a logarithmic function.
Let's note the key features of a logarithmic function. The shape will always
be similar to the one shown in the figure. The domain is \((0,\infty)\). The range is
\((-\infty,\infty)\). Additionally, the point \((1,0)\) is always included on a logarithmic
graph. Why does that make sense? What number do we have to raise any base by
to obtain 1? Any number raised to the zeroth power is always 1; therefore,
the point \((1,0)\) will always be on the graph of a logarithm.
We have been exploring a specific case of a logarithm, namely when the base
of the logarithm is 2. Consider the general form of a logarithmic function.
\(f(x)=log_ax\)
Investigate how \(a\) effects the shape of the graph using the applet below.
What do you notice? The larger the value of \(a\), the closer the logarithmic
function stays to the \(x\)-axis. Note that the significant features described
above hold true for all logarithmic functions (Kenney & Kastberg, 2013).
Domain: \((0,\infty)\)
Range: \((-\infty,\infty)\)
Significant Point: \((1,0)\)
With an understanding of logarithmic functions, students are better able to
see logarithms as their own entities with their own properties. This allows
them to reason when asked to take logarithms of negative numbers or explain
how logarithms yield outputs.
Logarithms as Inverse Functions
Students also struggle to see logarithms as inverse functions (Kenney &
Kastberg, 2013). They understand that logarithms and exponents are closely
related but lack an understanding of where that relationship comes from or
why it is significant. The following section outlines a way we could help
students make this connection.
Logarithms are the inverse of exponents. What does this mean? The
definition of an inverse function states that \(g(x)\) is the inverse of
\(f(x)\) if \(f(g(x))=x\). Consider the general equations for both
logarithmic and exponential functions.
\(f(x)=log_ax\)
\(g(x)=a^x\)
If we compose these two functions, it yields,
\(f(g(x))=log_a(a^x)\).
Going back to our fundamental understanding of logarithms, this statement
asks what is the exponent required of a to produce \(a^x\)? That value is
\(x\). Therefore, \(f(g(x))=x\), and we have shown that exponents are the
inverse of logarithmic functions. But why is this significant? One
researcher states, "Recognising the relationship that results from the
composition of inverse functions lead students to understand the
logarithmic function as a key tool for solving exponential
equations" (Kenney & Kastberg, 2013).
Let's consider an example. Suppose we are presented a problem about
exponential growth. The amount of bacteria left after a certain time can be
modeled by \(A=100e^{0.251t}\). Suppose there are 450 bacteria when we
investigate the sample. How long has it been decaying? The equation becomes
\(450=100e^{0.251t} \Rightarrow \frac{450}{100}=e^{0.251t}\)
Knowing that a logarithm is the inverse of an exponent is the key to
solving this question.
To see why, let's make a connection to algebra. Given the equation
\(2x=4\), how could we solve for \(x\)? Naturally, we think to divide by 2.
Why is that? Because division is the inverse operation of multiplication.
We often use the inverse to solve equations even if we do not call it that.
Back to the problem at hand. The x is in the exponent, so we need the
inverse of exponents, which is a logarithm. We can rewrite the equation.
\(log_e4.5=log_e(e^{0.251t})=0.251t \)
\(t=\frac{log_e4.5}{0.251} = 5.99\)
Without a logarithm, it would have been difficult to find the solution to
this equation.
Note that \(log_ex\) is often written as \(lnx\).
Students need to understand the connection between exponents and logarithms.
However, it is important they first develop a fundamental understanding of
a logarithm as a number and its own function. Then, students need to be
presented the inverse relationship in such a way that does not cause student
to think logarithms just disappear by putting them in exponent form (Kenney
& Kastberg, 2013).
Solving Equations with Logarithms
Studies have found that students struggle to manipulate logarithms
correctly. Often, they try to extend their work with polynomials to
logarithms, which leads to invalid manipulation and solutions
(Kenney & Kastberg, 2013). Traditionally, students are presented with the
logarithmic properties but are not given enough opportunities to
conceptualize and internalize the rules. Instead, they memorize the
properties or fall back on algebra techniques.
There are several ways students try to extend algebra to logarithms. First,
they treat the "\(log\)" in the equation as a variable. When presented with an
equation such as \(log(2x+3)=log(3x)\), student try to divide both sides by
"\(log\)." Second, students try to treat the "\(log\)" as its own entity
that can be distributed. For example, students will mistakenly believe
\(log(x+2)=logx+log2\) (Kenney & Kastberg, 2013). Third, students try to
"cancel" logarithms out of equations. To them, the statement
\(\frac{log100}{log10}=\frac{100}{10}=10\) because the logs "cancel out"
(Liang & Wood, 2005).
When teaching logarithms, it is important teachers understand these common
misconceptions and present logarithmic properties and equations in a way
that fosters deep understanding. Students need to be presented with
opportunities to convince themselves the logarithmic properties hold true
and to make correct connections with previously learned algebraic
manipulations. The following section presents a way students could be taught
about solving logarithmic equations.
Up to this point, we have dealt with equations dealing with only one
logarithmic statement. What happens when there are several logarithms such
as \(log_25+log_26.4\)? We know this statement represents a number, but is
there a way to combine these logarithms to make them easier to use? Luckily
for us, there is a way.
There are two important logarithm properties that will allow us to solve
problems involving several logarithms. (See Beyond the Classroom
for proof of each property listed below)
1. \(log_ab+log_ac=log_abc\)
2. \(log_ab-log_ac=log_a\frac{b}{c}\)
There is a third properties that helps us simplify statements where
exponents are in the logarithm.
3. \(log_ab^c=clog_a b\)
Does it seem strange that addition can be switched to addition? This may be
an unfamiliar technique but a powerful one (see History).
Let's convince ourselves this property holds true with an example.
The first logarithm property allows us to state \(log_25+log_26.4=log_232\).
We know the value of the logarithm on the right side. It is the exponent
required of 2 to obtain 32 or the number 5. Does \(log_25+log_26.4\) really
equal 5? Use a calculator,
to calculate that value. Amazing! This does hold true.
Doing a few examples of each logarithmic properties allows students to see
for themselves that these properties are true. With a deep belief in these
logarithmic properties and a fundamental understanding of logarithms as a
number, students will be less inclined to make the algebra mistakes
mentioned above.