Overview
A function can have many different representations. We often need to use these different representations in different settings. For example, polynomials usually are written as a sum of terms. Multiplication and division of polynomials usually work best in this representation. However, to find the roots or zeros of a polynomial, a factored representation is preferred. In order to change representations, we need to understand when two formulas represent the same function.
This section reviews the elementary rules of algebra and explores how these rules lead to algebraic properties of functions. The fundamental properties of addition and multiplication motivate additional rules, such as rules for fractions and powers. We will then apply these rules to understand properties of the exponential and logarithm functions. In order to learn to apply these rules, it is essential to recognize the structure of expressions as discussed in Subsection 2.2.2.
Subsection2.4.1Algebraic Properties
An algebraic representation of a function is an expression (formula) that computes the value of the dependent variable in terms of the independent variable. We say that two representations are equivalent if they give the same values. We will often need to be able to find equivalent expressions that will allow us to simplify our work. To find these equivalent representations, we rely on algebraic properties of the operations and elementary functions of algebra.
The properties of arithmetic form the foundation for all other equivalence properties. The two basic arithmetic operations are addition and multiplication. Subtraction and division are inverse operations and are ultimately defined in terms of addition and multiplication.
Properties of Addition
\(0\) is Additive Identity: \(x+0=x\)
Additive Inverses: Every number \(x\) has an additive inverse \(x\) so that \(x+x=0\text{,}\) namely \(\displaystyle x = 1 \cdot x\text{.}\)
Commutative Property: \(x+y = y+x\)
Associative Property: \(x+(y+z) = (x+y)+z\)
Subtraction is Addition: \(xy = x+y\)
Properties of Multiplication
\(1\) is Multiplicative Identity: \(x \cdot 1=x\)
Multiplicative Inverses: Every nonzero number \(x \ne 0\) has a multiplicative inverse \(\div x\) so that \(x \cdot \div x=1\text{,}\) namely \(\displaystyle \div x = \frac{1}{x}\) (reciprocal).
Commutative Property: \(x \cdot y = y \cdot x\)
Associative Property: \(x \cdot (y \cdot z) = (x \cdot y) \cdot z\)
Division is Multiplication: \(\displaystyle \frac{x}{y} = x \cdot \div y = x \cdot \frac{1}{y}\)

Multiplication Distributes Over Addition:
\begin{equation*}
x \cdot (y + z) = x \cdot y + x \cdot z
\end{equation*}
Example2.4.1
The expressions \(3(x4)+2\) and \(3x10\) are mathematically equivalent, even though they involve different operations. We show this by applying the properties in sequence.
\begin{align*}
3(x4)+2 &= (3x12)+2 & \text{Distributive}\\
&= (3x+12)+2 & \text{Additive Inverse}\\
&= 3x+(12+2) & \text{Associative}\\
&= 3x+10 & \text{Arithmetic}\\
&= 3x10 & \text{Subtraction}
\end{align*}
Students often encounter difficulties with expressions involving fractions. However, all of the algebra rules for fractions really are consequences of the basic properties of multiplication. Recognizing these connections and using them to guide your work will help you avoid common pitfalls. The first key is to remember that fractions are division and therefore really multiplication.
Theorem2.4.2Properties of Fractions
Divisors in the Denominator. \((\div a) \cdot (\div b) = \div(ab)\text{,}\) or \(\displaystyle \frac{1}{a} \cdot \frac{1}{b} = \frac{1}{ab}\)
Multiplication. \(\displaystyle a \cdot \frac{b}{c} = \frac{ab}{c}\) and \(\displaystyle \frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}\)
Reciprocals. \(\displaystyle \div \frac{a}{b} = \frac{b}{a}\)
Division. \(\displaystyle (\frac{a}{b}) \div (\frac{c}{d}) = \frac{ad}{bc}\)
Cancel Common Factors. \(\displaystyle \frac{ac}{bc} = \frac{a}{b}\) for \(c \ne 0\)
Addition with Common Denominator. \(\displaystyle \frac{a}{b} + \frac{c}{d} = \frac{ad+bc}{bd}\)
Each property results from the basic properties of addition. The following calculations emphasize these connections.

To show \(\div(ab) = (\div a) \cdot (\div b)\text{,}\) we need to show that the property of a multiplicative inverse is satisfied. This requires the commutative and associative properties.
\begin{equation*}
(ab) \cdot (\div a \cdot \div b) = (a \div a) \cdot (b \div b) = 1 \cdot 1 = 1\text{.}
\end{equation*}
Written in terms of fractions, this means \(\displaystyle \frac{1}{a} \cdot \frac{1}{b} = \frac{1}{ab}\text{.}\)

Multiplication is associative and commutative.
\begin{align*}
a \cdot \frac{b}{c} &= a (b \div c) = (a b) \div c = \frac{ab}{c}\\
\frac{a}{b} \cdot \frac{c}{d} &= (a {\div b}) \cdot (c {\div d}) = (ac) \cdot ({\div b} {\div d}) = (ac) {\div(bd)} = \frac{ac}{bd}
\end{align*}

Division is multiplication by inverse.
\begin{equation*}
(\frac{a}{b}) \div (\frac{c}{d}) = \frac{a}{b} \cdot \frac{d}{c} = \frac{ad}{bc}
\end{equation*}

Adding fractions applies the distributive property using the common denominator as the common factor.
\begin{align*}
\frac{a}{b} + \frac{c}{d} &= \frac{ad}{bd} + \frac{bc}{bd} = (ad) \,{\div(bd)} + (bc)\,{\div(bd)}\\
& = (ad+bc)\,{\div(bd)} = \frac{ad+bc}{bd}
\end{align*}
The basic properties of powers, shown below, can not be proved easily for all real numbers. We can motivate each property with integer powers using the properties of multiplication. Rational powers are more complicated and involve roots, but irrational powers are too difficult for this text.
Properties of Powers
Zero Power: \(b^0 = 1\) for \(b \ne 0\)
Inverse Power: \(\displaystyle b^{x} = \frac{1}{b^x}\)
Product with Common Base: \(b^x \, b^y = b^{x+y}\)
Quotient with Common Base: \(\displaystyle \frac{b^x}{b^y} = b^{xy}\)
Power of a power: \(\displaystyle (b^x)^y = b^{xy}\)
Product with Common Exponent: \(b^x \, c^x = (bc)^x\)
Quotient with Common Exponent: \(\displaystyle \frac{b^x}{c^x} = \left(\frac{b}{c}\right)^x\)
Example2.4.3
For integer exponents, a power means repeated multiplication (similar to how multiplication by an integer means repeated addition). So \(b^3 = b \cdot b \cdot b\text{.}\) The product properties are just about counting.
\begin{gather*}
b^3 \cdot b^2 = (bbb) \cdot (bb) = b^5 = b^{3+2}\\
(b^2)^3 = (bb)(bb)(bb) = b^{2 \cdot 3}\\
(ab)^3 = (ab)(ab)(ab) = (aaa)(bbb) = a^3 b^3
\end{gather*}
(Do you see the commutative and associative properties in action?)
The quotient properties for integer powers result from canceling common factors followed by counting.
\begin{gather*}
\frac{b^6}{b^2} = \frac{(bb)(bbbb)}{bb} = bbbb = b^{62}\\
\left(\frac{b}{c}\right)^3 = \frac{b}{c} \cdot \frac{b}{c} \cdot \frac{b}{c} = \frac{b^3}{c^3}
\end{gather*}
Addition and powers have no convenient properties. Many mistakes occur when students forget this and imagine that powers distribute like multiplication. (It doesn't!)
Example2.4.4
To illustrate that \((a+b)^2 \ne a^2+b^2\text{,}\) consider the numbers \(a=2\) and \(b=3\text{.}\) The first expression gives
\begin{equation*}
(a+b)^2 = (2+3)^2 = 5^2 = 25
\end{equation*}
while the second expression gives
\begin{equation*}
a^2+b^2=2^2+3^2=4+9=13.
\end{equation*}
The proper way to expand the first expression is to think of the power as repeated multiplication and apply the distributive property. This is often called the FOIL method:
\begin{equation*}
(a+b)^2=(a+b)(a+b)=a^2+2ab+b^2\text{.}
\end{equation*}
Example2.4.5
To illustrate that \(\sqrt{a^2+b^2} \ne a+b\text{,}\) again consider the numbers \(a=2\) and \(b=3\text{.}\) The first expression gives
\begin{equation*}
\sqrt{a^2+b^2} = \sqrt{2^2+3^2}=\sqrt{13},
\end{equation*}
and the second expression gives
\begin{equation*}
a+b=2+3=5.
\end{equation*}
There is no convenient way to expand a square root.
If you see any operation acting on a sum (e.g., powers, roots or functions), unless you know a specific rule, you should not break up the formula into individual parts. Such rules are called sum rules and will play a very important role in calculus.
Subsection2.4.2Exponential and Logarithm Properties
The elementary exponential functions \(\exp_b(x)=b^x\) provide an interesting exception to the general principle that a sum inside an operation can not be simply rewritten. They inherit a property from powers:
\begin{equation*}
\exp_b(x+y) = b^{x+y} = b^x \cdot b^y\text{.}
\end{equation*}
Such a rule might be remembered as “the exponential of a sum is the product of exponentials”. Similarly, we have related properties for quotients and powers.
Properties of Elementary Exponentials
 Exponential Sum Rule: \(\exp_b(x+y) = \exp_b(x) \cdot \exp_b(y)\) or \(b^{x+y} = b^x b^y\)
 Exponential Difference Rule: \(\exp_b(xy) = \exp_b(x) \div \exp_b(y)\) or \(\displaystyle b^{xy} = \frac{b^x}{b^y}\)
 Exponential of a Product: \(\displaystyle b^{xy} = (b^x)^y\) and \(b^{xy} = (b^y)^x\)
The logarithm with base \(b\) is defined as the inverse function of the corresponding elementary exponential function. Thus \(x=\log_b(u)\) is defined as the solving to the equation \(b^x=u\) for \(x\text{.}\) For each property of the exponential, the logarithm has a related, inverse property.
Properties of Logarithms
Inverses Cancel: \(\log_b(b^x) = x\) (all \(x\)) and \(b^{\log_b(x)} = x\) (\(x \gt 0\))
Product: \(\log_b(u \cdot v)= \log_b(u) + \log_b(v) \)
Quotient: \(\displaystyle \log_b\left(\frac{u}{v}\right)= \log_b(u)  \log_b(v) \)
Power: \(\log_b(u^p) = p \cdot \log_b(u)\)
Historically, the logarithm was invented so that multiplication and division calculations could be solved using the much simpler operations of addition and subtraction. Engineers and scientists would often reference logarithm tables. The slide rule was a mechanical implementation, where the lengths corresponding to logarithm values were added physically to get a new length.
Example2.4.6
A reference table shows \(\log(2) \approx 0.30103\text{,}\) \(\log(3) \approx 0.47712\) and \(\log(5) \approx 0.69897\text{.}\) Use properties of logarithms to determine \(\log(1.2)\text{.}\)
SolutionStart by writing 1.2 as a product of the factors 2, 3, and 5.
\begin{equation*}
1.2 = \frac{12}{10} = \frac{4(3)}{2(5)} = \frac{2(3)}{5}
\end{equation*}
The properties of logarithms allow us to rewrite \(\log(1.2)\) as
\begin{align*}
\log(1.2) &= \log\left(\frac{2(3)}{5}\right) \\
&= \log(2(3))  \log(5) & \text{(Quotient)} \\
&= \log(2) + \log(3)  \log(5) & \text{(Product)} \\
&\approx 0.30103 + 0.47712  0.69897 = 0.07918.
\end{align*}
We still use logarithms in order to expand formulas expressed as products and powers in terms of sums and products.
Example2.4.7
Expand \(\displaystyle \log\left(\frac{4x^3\sqrt{2x+5}}{(x^2+3)^5}\right)\) as far as possible.
SolutionEach factor of the expression inside the logarithm will get its own term using the product and quotient rules for logarithms. If we think of every division in terms of negative powers, then we only need to deal with products. In particular, the inner expression can be rewritten
\begin{equation*}
\frac{4x^3\sqrt{2x+5}}{(x^2+3)^5} = 4x^3(2x+5)^{\frac{1}{2}}(x^2+3)^{5}\text{.}
\end{equation*}
The factors identified are \(4\text{,}\) \(x^3\text{,}\) \((2x+5)^{\frac{1}{2}}\text{,}\) and \((x^2+3)^{5}\text{.}\) Using the logarithm's product rule followed by the logarithm's power rule, we find
\begin{align*}
\log\left(\frac{4x^3\sqrt{2x+5}}{(x^2+3)^5}\right) &= \log(4) + \log(x^3) + \log\left( (2x+5)^{\frac{1}{2}}\right) + \log\left( (x^2+3)^{5} \right) \\
&= \log(4) + 3\log(x) + {\textstyle \frac{1}{2}}\log(2x+5) 5 \log(x^2+3).
\end{align*}
Notice that we could have used the quotient rule of logarithms instead of negative powers to get the term \(5 \log(x^2+3)\text{.}\)
The properties of logarithms help solve us equations, particularly where the variable is in an exponent. Because inverse functions are onetoone, we can apply such a function to both sides of an equation with positive values and form a new equation that is equivalent to the first. The properties of the logarithm can then be used to our advantage.
Example2.4.8
Solve the equation \(3 \cdot 2^{3x} = 5\) using the logarithm base 10.
SolutionIt is usually best to isolate the exponential term first.
\begin{gather*}
3 \cdot 2^{3x} = 5 \\
2^{3x} = \frac{5}{3}
\end{gather*}
We next apply the logarithm base 10 to both sides of this equation, which then allows us to apply the logarithm power rule on the left. Then we can isolate \(x\text{.}\)
\begin{gather*}
\log_{10}(2^{3x}) = \log_{10}({\textstyle \frac{5}{3}}) \\
3x \cdot \log_{10}(2) = \log_{10}({\textstyle \frac{5}{3}}) \\
x = \frac{\log_{10}({\textstyle \frac{5}{3}})}{3 \log_{10}(2)}
\end{gather*}
Alternatively, we could have applied the logarithm at the very first. This would require using the logarithm product rule on the left.
\begin{gather*}
\log_{10}(3 \cdot 2^{3x}) = \log_{10}(5) \\
\log_{10}(3) + \log_{10}(2^{3x}) = \log_{10}(5) \\
\log_{10}(2^{3x}) = \log_{10}(5)  \log_{10}(3) \\
3x \log_{10}(2) = \log_{10}(5)  \log_{10}(3) \\
x = \frac{\log_{10}(5)  \log_{10}(3)}{3\log_{10}(2)}
\end{gather*}
The properties of logarithms allow us to compute logarithms with uncommon bases using logarithms that we know using the change of base formula.
Theorem2.4.9
For any two exponential bases \(b\) and \(c\text{,}\)
\begin{equation*}
\log_b(u) = \frac{\log_c(u)}{\log_c(b)}\text{.}
\end{equation*}
Consider \(x=\log_b(u)\text{,}\) which solves \(b^x=u\text{.}\) If we solve this equation using \(\log_c\) to both sides, we find the change of base formula.
\begin{gather*}
b^x = u \\
\log_c(b^x) = \log_c(u) \\
x \log_c(b) = \log_c(u) \\
x = \log_b(u) = \frac{\log_c(u)}{\log_c(b)}.
\end{gather*}
Closely related to change of base formula is the fact that we can rewrite any power with a positive base using a composition with an elementary exponential. We will soon discover that the number \(e\) is the natural exponential base. Thus, every power can be rewritten in terms of the natural exponential function. The logarithm with this base is the natural logarithm \(\ln\text{.}\)
Example2.4.10
Rewrite \(f(x) = 4 \cdot 3^{2x}\) using the natural exponential function.
SolutionOne approach is to use the inverse property, \(\exp(\ln(u)) = e^{\ln(u)} = u\text{,}\) on the factor with the power. Then use logarithm properties to simplify (expand) the power.
\begin{align*}
f(x) &= 4 \cdot 3^{2x} \\
&= 4 \cdot e^{\ln(3^{2x})} & \text{(Inverse)} \\
&= 4 \cdot e^{2x \ln(3)} & \text{(Power)} \\
&= 4 \cdot e^{2 \ln(3) \, x} & \text{(Commute)}
\end{align*}
Another approach is to just rewrite the base using the inverse property, \(3 = e^{\ln(3)}\text{,}\) and then finish by using properties of powers.
\begin{align*}
f(x) &= 4 \cdot 3^{2x} \\
&= 4 \cdot (e^{\ln(3)})^{2x} & \text{(Inverse)} \\
&= 4 \cdot e^{2x \ln(3)} & \text{(Power of Product)} \\
&= 4 \cdot e^{2 \ln(3) \, x} & \text{(Commute)}
\end{align*}
We use base \(e\) for exponentials so much that we summarize the statement as a theorem.
Theorem2.4.11
Every exponential function \(f(x) = A b^x\) can be written using the natural exponential \(f(x) = Ae^{kx}\) where \(k = \ln(b)\text{.}\)
Example2.4.12
A population \(P\) is an exponential function of time \(t\text{,}\) \(P = Ae^{kt}\text{.}\) Suppose that \(P=500\) when \(t=0\) and the population triples every 5 years, find the formula for \(P\text{.}\)
SolutionThis is a parametrized model \(P=Ae^{kt}\text{.}\) We use the data \((t,P)=(0,500)\) and \((t,P)=(5,1500)\) (population triples) to create equations based on our model.
\begin{equation*}
\left\{ \begin{matrix} 500 = Ae^{0k} \\ 1500 = Ae^{5k} \end{matrix} \right.
\end{equation*}
The first equation gives \(A=500\text{.}\) Substituting that into the second equation, we can solve for \(k\text{.}\)
\begin{gather*}
Ae^{5k} = 1500 \\
500 e^{5k} = 1500 \\
e^{5k} = 3 \\
5k = \ln(3) \\
k = {\textstyle \frac{1}{5}} \ln(3)
\end{gather*}
Using these parameters, we have our model
\begin{equation*}
P = 500 e^{\frac{1}{5} \ln(3) t}\text{.}
\end{equation*}
1
Show using the elementary properties of addition and multiplication why \(2(x+3)1=2x+5\text{.}\)
2
Show using the elementary properties of addition and multiplication why \((x+3)(x1)=x^2+2x3\text{.}\)
3
A student made a mistake writing \(\displaystyle \frac{3x+1}{x} = \frac{3+1}{1} = 4\text{.}\) What did the student do? Why was it incorrect?
4
A student made a mistake writing \(\displaystyle x \cdot \frac{2x+1}{x+3} = \frac{2x^2+x}{x^2+3x}\text{.}\) What did the student do? Why was it incorrect?
5
A student made a mistake writing \(\displaystyle 3 \cdot 2^x = 6^x\text{.}\) What did the student do? Why was it incorrect?
6
For an unknown base \(b\text{,}\) we have \(\log_b(2)=0.3\text{.}\) Use the rules of logarithms to find each of the following.
 \(\log_b(8)\)
 \(\log_b(\frac{1}{\sqrt{2}})\)
 \(\log_b(4b^2)\)
Can you identify the value \(b\text{?}\)
7
For an unknown base \(b\text{,}\) we have \(\log_b(2) \approx 0.3562\) and \(\log_b(3) \approx 0.5646\text{.}\) Use the rules of logarithms to find each of the following.
 \(\log_b(6)\)
 \(\log_b(72)\)
 \(\log_b(\frac{4}{9})\)
8
Expand \(\displaystyle \log\left( 3x^5(2x+1)^4 \right)\) as far as possible.
9
Expand \(\displaystyle \log\left( \frac{(x^2+4)^3}{x^4(3x+1)} \right)\) as far as possible.
10
Expand \(\displaystyle \log\left( \sqrt{5(x^2+1)} \right)\) as far as possible.
11
Rewrite the expanded formula \(\displaystyle 2 + 3\log(x)  \log(2x+1)\) as the logarithm (base 10) of a single expression.
12
Rewrite the expanded formula \(\displaystyle {\textstyle \frac{1}{2}}\ln(x)  \ln(x+1)  \ln(x1)\) as the logarithm (base e) of a single expression.
13
Use the natural logarithm to solve the equation \(4 \cdot 5^x = 3\text{.}\)
14
Use the natural logarithm to solve the equation \(2 \cdot 3^{x} = 3 \cdot 2^{x}\text{.}\)
15
Write the function \(f(x) = 4^x\) using an exponential with base \(e\text{.}\)
16
Write the function \(f(x) = 5 \cdot 0.25^x\) using an exponential with base \(e\text{.}\)
17
Write the function \(f(x) = x^4\) using an exponential with base \(e\) for \(x \gt 0\text{.}\)
18
Write the function \(f(x) = x^{2x}\) using an exponential with base \(e\) for \(x \gt 0\text{.}\)
19
Find an exponential model \(y=Ae^{kx}\) satisfying the states \((x,y)=(0,3)\) and \((x,y)=(5,9)\text{.}\)
20
Find an exponential model \(y=Ae^{kx}\) satisfying the states \((x,y)=(1,3)\) and \((x,y)=(4,6)\text{.}\)
21
Chemistry texts often give a formula for the mass \(M\) of radioactively decaying element in terms of the halflife,
\begin{equation*}
M=M_0 e^{\ln(2)t/\tau_{\frac12}}\text{,}
\end{equation*}
where \(M_0\) is the mass at time \(t=0\) and \(\tau_{\frac12}\) is the halflife. Use the properties of exponentials and logarithms to show that this is equivalent to
\begin{equation*}
M = M_0 \cdot \Big(\frac{1}{2}\Big)^{t/\tau_{\frac12}}\text{.}
\end{equation*}
22
The isotope of plutonium Pu239 has a halflife of 24,110 years. For an initial mass of 1 kg, how much plutonium remains after 100 years?
23
A population that doubles in size every 5 years currently has 1000 individuals. What will the population be in 4 years?