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In Section LogarithmicFunctions, we introduced the logarithmic functions as inverses of exponential functions and discussed a few of their
functional properties from that perspective. In this section, we explore the algebraic properties of logarithms. Historically,
these have played a huge role in the scientific development of our society since, among other things, they were used to
develop analog computing devices called slide rules which enabled scientists and engineers to perform accurate calculations
leading to such things as space travel and the moon landing.
As we shall see shortly, logs inherit analogs of all of the properties of exponents you learned in Elementary and Intermediate
Algebra. We first extract two properties from Theorem logfcnprops to remind us of the definition of a logarithm as the inverse of an
exponential function.
(Inverse Properties of Exponential and Logarithmic Functions)
Let \(b > 0\), \(b \neq 1\).
\(b^{a} = c\) if and only if \(\log _{b}(c) = a\). That is, \(\log _{b}(c)\) is the exponent you put on \(b\) to obtain \(c\).
\(\log _{b} \left (b^{x}\right ) = x\) for all \(x\) and \(b^{\log _{b}(x)} = x\) for all \(x > 0\)
Next, we spell out what it means for exponential and logarithmic functions to be one-to-one.
(One-to-one Properties of Exponential and Logarithmic Functions)
Let \(f(x) = b^{x}\) and \(g(x) = \log _{b}(x)\) where \(b>0\), \(b\neq 1\). Then \(f\) and \(g\) are one-to-one and
\(b^{u} = b^{w}\) if and only if \(u=w\) for all real numbers \(u\) and \(w\).
\(\log _{b}(u) = \log _{b}(w)\) if and only if \(u=w\) for all real numbers \(u > 0\), \(w > 0\).
Next, we re-state Theorem algpropexpfcns for reference below.
(Algebraic Properties of Exponential Functions) Let \(f(x) = b^{x}\) be an exponential function (\(b > 0\), \(b\neq 1\)) and let \(u\) and \(w\) be real
numbers.
Product Rule:\(f(u+w) = f(u) f(w)\). In other words, \(b^{u+w} = b^{u} b^{w}\)
Quotient Rule:\(f(u-w) = \frac {f(u)}{f(w)}\). In other words, \(b^{u-w} = \frac {b^{u}}{b^{w}}\)
Power Rule:\(\left (f(u)\right )^w = f(uw)\). In other words, \(\left (b^{u}\right )^{w} = b^{uw}\)
To each of these properties of listed in Theorem algpropexpfcns, there corresponds an analogous property of logarithmic functions. We list
these below in our next theorem.
(Algebraic Properties of Logarithmic Functions) Let \(g(x) =\log _{b}(x)\) be a logarithmic function (\(b > 0\), \(b\neq 1\)) and let \(u>0\) and \(w>0\) be real
numbers.
Product Rule:\(g(uw) = g(u)+ g(w)\). In other words, \(\log _{b}(uw) = \log _{b}(u) + \log _{b}(w)\)
Power Rule:\(g\left (u^{w}\right ) =w g(u)\). In other words, \(\log _{b}\left (u^{w}\right ) = w \log _{b}(u)\)
There are a couple of different ways to understand why Theorem algproplogfcns is true. For instance, consider the product rule:
\(\log _{b}(uw) = \log _{b}(u) + \log _{b}(w)\).
Let \(a = \log _{b}(uw)\), \(c = \log _{b}(u)\), and \(d = \log _{b}(w)\). Then, by definition, \(b^{a} = uw\), \(b^{c} = u\) and \(b^{d} = w\). Hence, \(b^{a} = uw = b^{c} b^{d} = b^{c+d}\), so that \(b^{a} = b^{c+d}\).
By the one-to-one property of \(b^{x}\), \(b^{a} = b^{c+d}\) gives \(a = c+d\). In other words, \(\log _{b}(uw) = \log _{b}(u) + \log _{b}(w)\). The remaining properties are proved similarly.
From a purely functional approach, we can see the properties in Theorem algproplogfcns as an example of how inverse functions
interchange the roles of inputs in outputs.
For instance, the Product Rule for exponential functions given in Theorem algpropexpfcns, \(f(u+w) = f(u)f(w)\), says that adding inputs results in multiplying
outputs.
Hence, whatever \(f^{-1}\) is, it must take the products of outputs from \(f\) and return them to the sum of their respective inputs. Since the
outputs from \(f\) are the inputs to \(f^{-1}\) and vice-versa, we have that that \(f^{-1}\) must take products of its inputs to the sum
of their respective outputs. This is precisely one way to interpret the Product Rule for Logarithmic functions:
\(g(uw) = g(u) + g(w)\).
The reader is encouraged to view the remaining properties listed in Theorem algproplogfcns similarly.
The following examples help build familiarity with these properties. In our first example, we are asked to ‘expand’
the logarithms. This means that we read the properties in Theorem algproplogfcns from left to right and rewrite products
inside the log as sums outside the log, quotients inside the log as differences outside the log, and powers
inside the log as factors outside the log. (Interestingly enough, it is the exact opposite process (which
we will practice later) that is most useful in Algebra, the utility of expanding logarithms becomes apparent in
Calculus.)
Expand the following using the properties of logarithms and simplify. Assume when necessary that all quantities represent
positive real numbers.
\(\log _{2}\left (\frac {8}{x}\right )\)
\(\log _{0.1} \left (10 x^2 \right )\)
\(\ln \left (\frac {3}{et}\right )^2\)
\(\log \sqrt [3]{\frac {100 x^2}{yz^5}}\)
\(\log _{117}\left (x^2 - 4\right )\)
To expand \(\log _{2}\left (\frac {8}{x}\right )\), we use the Quotient Rule identifying \(u = 8\) and \(w=x\) and simplify.
In the expression \(\log _{0.1} \left (10 x^2 \right )\), we have a power (the \(x^2\)) and a product, and the question becomes which property, Power
Rule or Product Rule to use first.
In order to use the Power Rule, the entire quantity inside the log must be raised to the same exponent. Since
the exponent \(2\) applies only to the \(x\), we first apply the Product Rule with \(u=10\) and \(w=x^2\). Once the \(x^2\) is by itself inside the
log, we apply the Power Rule with \(u=x\) and \(w=2\).
We have a power, quotient and product occurring in \(\ln \left (\frac {3}{et}\right )^2\). Since the exponent \(2\) applies to the entire quantity inside
the logarithm, we begin with the Power Rule with \(u=\frac {3}{et}\) and \(w = 2\).
Next, we see the Quotient Rule is applicable, with \(u=3\) and \(w=et\), so we replace \(\ln \left (\frac {3}{et}\right )\) with the quantity \(\ln (3) - \ln (et)\).
Since \(\ln \left (\frac {3}{et}\right )\) is being multiplied by \(2\), the entire quantity \(\ln (3) - \ln (et)\) is multiplied by \(2\).
Finally, we apply the Product Rule with \(u=e\) and \(w=x\), and replace \(\ln (et)\) with the quantity \(\ln (e) + \ln (t)\), and simplify, keeping in mind that
the natural log is log base \(e\).
In Theorem algproplogfcns, there is no mention of how to deal with radicals. However, thinking back to Definition rationalexponentdefn, we can
rewrite the cube root as a \(\frac {1}{3}\) exponent. We begin by using the Power Rule (At this point in the text, the
reader is encouraged to carefully read through each step and think of which quantity is playing the role of \(u\) and
which is playing the role of \(w\) as we apply each property.) , and we keep in mind that the common log is log base
\(10\).
At first it seems as if we have no means of simplifying \(\log _{117}\left (x^2-4\right )\), since none of the properties of logs addresses the
issue of expanding a difference inside the logarithm. However, we may factor \(x^2 - 4 = (x+2)(x-2)\) thereby introducing a product
which gives us license to use the Product Rule. Assuming both \(x+2>0\) and \(x-2>0\), that is, \(x >2\) we expand as follows.
A couple of remarks about Example expandlogex are in order. First, if we take a step back and look at each problem in the foregoing
example, a general rule of thumb to determine which log property to apply first when faced with a multi-step problem is to
apply the logarithm properties in the ‘reverse order of operations.’
For example, if we were to substitute a number for \(x\) into the expression \(\log _{0.1} \left (10 x^2 \right )\), we would first square the \(x\), then multiply by \(10\). The last
step is the multiplication, which tells us the first log property to apply is the Product Rule. The last property of logarithm to
apply would the be the power rule applied to \(\log _{0.1}(x^2)\).
Second, the equivalence \(\log _{117}\left (x^2-4\right ) = \log _{117}(x+2) + \log _{117}(x-2)\) is valid only if \(x > 2\). Indeed, the functions \(f(x) = \log _{117}\left (x^2-4\right )\) and \(g(x) = \log _{117}(x+2) + \log _{117}(x-2)\) have different domains, and, hence, are
different functions. (We leave it to the reader to verify the domain of \(f\) is \((-\infty , -2) \cup (2,\infty )\) whereas the domain of \(g\) is \((2,\infty )\).) In
general, when using log properties to expand a logarithm, we may very well be restricting the domain as we do
so.
One last comment before we move to reassembling logs from their various bits and pieces. The authors are well aware of the
propensity for some students to become overexcited and invent their own properties of logs like \(\log _{117}\left (x^2-4\right ) = \log _{117}\left (x^2\right ) - \log _{117}(4)\), which simply isn’t true, in
general. The unwritten (The authors relish the irony involved in writing what follows.) property of logarithms is that if it
isn’t written in a textbook, it probably isn’t true.
Use the properties of logarithms to write the following as a single logarithm.
\(\log _{3}(x-1) - \log _{3}(x+1)\)
\(\log (x) + 2\log (y) - \log (z)\)
\(4\log _{2}(x) + 3\)
\(-\ln (t) - \frac {1}{2}\)
Whereas in Example expandlogex we read the properties in Theorem algproplogfcns from left to right to expand logarithms, in this example we read them
from right to left.
The difference of logarithms requires the Quotient Rule: \(\log _{3}(x-1) - \log _{3}(x+1) = \log _{3}\left (\frac {x-1}{x+1}\right )\).
In the expression, \(\log (x) + 2\log (y) - \log (z)\), we have both a sum and difference of logarithms.
Before we use the product rule to combine \(\log (x) + 2\log (y)\), we note that we need to apply the Power Rule to rewrite the
coefficient \(2\) as the power on \(y\). We then apply the Product and Quotient Rules as we move from left to right.
We begin rewriting \(4\log _{2}(x) + 3\) by applying the Power Rule: \(4\log _{2}(x) = \log _{2}\left (x^4\right )\).
In order to continue, we need to rewrite \(3\) as a logarithm base \(2\). From Theorem invpropslogs, we know \(3 = \log _{2}\left (2^3\right )\). Rewriting \(3\) this way
paves the way to use the Product Rule.
To get started with \(-\ln (t) - \frac {1}{2}\), we rewrite \(-\ln (t)\) as \((-1) \ln (t)\). We can then use the Power Rule to obtain \((-1)\ln (t) = \ln \left (t^{-1}\right )\).
As in the previous problem, in order to continue, we need to rewrite \(\frac {1}{2}\) as a natural logarithm. Theorem invpropslogs gives us
\(\frac {1}{2} = \ln \left (e^{1/2}\right ) = \ln \left (\sqrt {e}\right )\). Hence,
As we would expect, the rule of thumb for re-assembling logarithms is the opposite of what it was for dismantling them. That
is, to rewrite an expression as a single logarithm, we apply log properties following the usual order of operations: first, rewrite
coefficients of logs as powers using the Power Rule, then rewrite addition and subtraction using the Product and Quotient
Rules, respectively, as written from left to right.
Additionally, we find that using log properties in this fashion can increase the domain of the expression. For example, we leave
it to the reader to verify the domain of \(f(x) = \log _{3}(x-1) - \log _{3}(x+1)\) is \((1,\infty )\) but the domain of \(g(x) = \log _{3}\left (\frac {x-1}{x+1}\right )\) is \((-\infty , -1) \cup (1, \infty )\). We’ll need to keep this in mind in Section LogarithmicEquationsandInequalities since such
manipulations can result in extraneous solutions.
The two logarithm buttons commonly found on calculators are the ‘LOG’ and ‘LN’ buttons which correspond to the common
and natural logs, respectively. Suppose we wanted an approximation to \(\log _{2}(7)\). The answer should be a little less than \(3\), (Can you
explain why?) but how do we coerce the calculator into telling us a more accurate answer? We need the following
theorem.
(Change of Base Formulas) Let \(a,b >0\), \(a,b \neq 1\).
\(a^{x} = b^{x \log _{b}(a)}\) for all real numbers \(x\).
\(\log _{a}(x) = \frac {\log _{b}(x)}{\log _{b}(a)}\) for all real numbers \(x > 0\).
To prove these formulas, consider \(b^{x \log _{b}(a)}\). Using the Power Rule, we can rewrite \(x \log _{b}(a)\) as \(\log _{b}\left (a^{x}\right )\). Following this with the Inverse Properties in
Theorem invpropslogs, we get
We get the result by dividing both sides of the equation \(\log _{a}(x) \cdot \log _{b}(a) = \log _{b}(x)\) by \(\log _{b}(a)\).
Of course, the authors can’t help but point out the inverse relationship between these two change of base formulas. To
change the base of an exponential expression, we multiply the input by the factor \(\log _{b}(a)\). To change the base of a logarithmic
expression, we divide the output by the factor \(\log _{b}(a)\).
While, in the grand scheme of things, both change of base formulas are really saying the same thing, the
logarithmic form is the one usually encountered in Algebra while the exponential form isn’t usually introduced until
Calculus.
Use an appropriate change of base formula to convert the following expressions to ones with the indicated base. Verify your
answers using a graphing utility, as appropriate.
\(3^{2}\) to base \(10\)
\(2^{x}\) to base \(e\)
\(\log _{4}(5)\) to base \(e\)
\(\ln (x)\) to base \(10\)
We apply the Change of Base formula with \(a=3\) and \(b=10\) to obtain \(3^2 = 10^{2 \log (3)}\). The reader is invited to check this answer using
your favorite graphing utility.
Here, \(a=2\) and \(b = e\) so we have \(2^{x} = e^{x \ln (2)}\). Using desmos, we find the graphs of \(f(x) = 2^x\) and \(g(x) = e^{x \ln (2)}\) appear to overlap perfectly.
Applying the change of base with \(a=4\) and \(b=e\) leads us to write \(\log _{4}(5) = \frac {\ln (5)}{\ln (4)}\). Evaluating this gives the numerical approximation
\(\frac {\ln (5)}{\ln (4)} \approx 1.16\).
To check our answer we know that, by definition, \(\log _{4}(5)\) is the exponent we put on \(4\) to get \(5\), so a number a little larger than \(1\)
seems reasonable.
Taking this one step further, we use a graphing utility and find \(4^{ \frac {\ln (5)}{\ln (4)}} = 5\), which means if the machine is lying to us about the first
answer it gave us, at least it is being consistent.
We write \(\ln (x) = \log _{e}(x) = \frac {\log (x)}{\log (e)}\). We graph both \(f(x) = \ln (x)\) and \(g(x) = \frac {\log (x)}{\log (e)}\) using desmos and find both graphs appear to be identical.
What Theorem changeofbase really tells us is that all exponential and logarithmic functions are just scalings of one another. Not only does
this explain why their graphs have similar shapes, but it also tells us that we could do all of mathematics with a single base, be
it \(10\), \(0.42\), \(\pi \), or \(117\).
As mentioned in Section ExponentialFunctions, the ‘natural’ base, base \(e\), features prominently in mathematical applications as we’ll see in Section ExpLogApplications.
Hence, we conclude this section by specifying Theorem changeofbase to this case.
Conversion to the Natural Base: Suppose \(b>0\), \(b \neq 1\). Then
\(b^{x} = e^{x \ln (b)}\) for all real numbers \(x\).
\(\log _{b}(x) = \frac {\ln (x)}{\ln (b)}\) for all real numbers \(x > 0\).
PreCalculus