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Section 2.4 Exponents, Inverses and Logarithms


We saw that geometric sequences involved powers of the common ratio. Powers play a very important role in many mathematical models. There are two major families of functions that involve powers: the power functions and the exponential functions.

Because a function defines a map from one variable to another variable, say \(x \mapsto y\text{,}\) we are often interested in the inverse relation, \(y \mapsto x\text{.}\) When the inverse relation defines a function it is called the inverse function. The inverse functions associated with power functions are roots; the inverse functions associated with exponential functions are logarithms.

In this section, we review the properties of exponents and characterize power and exponential functions. We discuss the concept of inverse functions and introduce roots and logarithms as inverse to elementary power and exponential function, respectively. We learn to use inverse functions to answer questions associated with reverse maps in sequences.

Subsection 2.4.1 Properties of Exponents

Consider the geometric sequence defined recursively by \(x_{n} = 3 x_{n-1}\) and initial value \(x_2 = 4\text{.}\) We know how to generate additional sequence values from the recursive equation:

\begin{gather*} x_3 = 3 x_2 = 3 \cdot 4 = 12 \\ x_4 = 3 x_3 = 3 \cdot 12 = 36 \\ x_5 = 3 x_4 = 3 \cdot 36 = 108 \\ \vdots \end{gather*}

We are used to multiplying out the values to get a simple final value. Look at what happens if we suspend the arithmetic facts and just write out the formulas.

\begin{align*} x_3 &= 3 x_2 = 3 \cdot 4 \\ x_4 &= 3 x_3 = 3 \cdot 3 \cdot 4 \\ x_5 &= 3 x_4 = 3 \cdot 3 \cdot 3 \cdot 4 \\ \vdots \end{align*}

Repeated multiplication corresponds to integer powers, so that we can identify a pattern.

\begin{align*} x_3 &= 3^1 \cdot 4 \\ x_4 &= 3^2 \cdot 4 \\ x_5 &= 3^3 \cdot 4 \\ \vdots \end{align*}

The power on the multiple 3 is always the index value minus 2,

\begin{equation*} x_n = 3^{(n-2)} \cdot 4, \end{equation*}

which is precisely what we would get using the explicit formula for geometric sequences

The basic properties of powers are all motivated by the idea that an integer power corresponds to repeated multiplication. Rational powers are more complicated and involve roots. Irrational powers are limits of rational powers—we will deal with them using exponentials and logarithms.

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: \(\displaystyle \frac{b^x}{b^y} = b^{x-y}\)

  • Power of a power: \(\displaystyle (b^x)^y = b^{xy}\)

  • Product with Common Exponent: \(b^x \, c^x = (bc)^x\)

  • Quotient with Common Exponent: \(\displaystyle \frac{b^x}{c^x} = \left(\frac{b}{c}\right)^x\)

We illustrate several of the properties in the context of integer powers. 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.

Example 2.4.1
\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*}

The quotient properties for integer powers result from simplifying fractions, canceling common factors when possible.

Example 2.4.2
\begin{gather*} \frac{b^6}{b^2} = \frac{(bb)(bbbb)}{bb} = bbbb = b^{6-2}\\ \left(\frac{b}{c}\right)^3 = \frac{b}{c} \cdot \frac{b}{c} \cdot \frac{b}{c} = \frac{b^3}{c^3} \end{gather*}

The zero power property is necessary for the power of a sum rule to remain consistent. We know that \(b^{x+0}=b^x\text{.}\) But the properties of powers also mean \(b^{x+0} = b^x \cdot b^0\text{.}\) For these to both be true requires \(b^x = b^x \cdot b^0\) or that \(b^0 = 1\text{.}\)

Addition and powers have no convenient properties. Many mistakes occur when students forget this and imagine that powers distribute like multiplication. (It doesn't!)

Example 2.4.3

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*}

Subsection 2.4.2 Exponential and Power Functions

A geometric sequence \(x\) with a constant multiple \(\rho\) and a given initial value \(x_k\) has an explicit formula

\begin{equation*} x_n = x_k \cdot \rho^{n-k}\text{.} \end{equation*}

Applying the properties of powers, this can be rewritten

\begin{equation*} x_n = x_k \cdot \frac{\rho^n}{\rho^k} = \frac{x_k}{\rho^k} \cdot \rho^n. \end{equation*}

Notice that the independent variable, the index \(n\text{,}\) appears in the power of the expression \(\rho^n\text{.}\) The other expression \(\displaystyle \frac{x_k}{\rho^k}\) depends only on the initial value. That is, every geometric sequence \(x\) can be written in the form

\begin{equation*} x_n = A \cdot \rho^n \end{equation*}

where \(A\) is a constant that depends on the initial value. We call such a function an exponential function.

Definition 2.4.4 Exponential Function

An exponential function with base \(b \gt 0\) and independent variable \(t\) is a function of the form

\begin{equation*} f(t) = A \cdot b^t\text{,} \end{equation*}

where \(A\) is a constant.

When the constant \(A=1\text{,}\) we call the function an elementary exponential function and write

\begin{equation*} \exp_b(t) = b^t\text{.} \end{equation*}

The distinguishing feature of an exponential function is that the base is a constant and the independent variable is the exponent. When the base of a power is the independent variable and the exponent is constant, we have a power function.

Definition 2.4.5 Power Function

A power function with power \(p\) and independent variable \(t\) is a function of the form

\begin{equation*} f(t) = A \cdot t^k\text{,} \end{equation*}

where \(A\) is a constant.

When the constant \(A=1\text{,}\) we call the function an elementary power function and write

\begin{equation*} \mathrm{pow}_p(t) = t^p\text{.} \end{equation*}

Where exponential functions naturally arise in geometric sequences, power functions often appear in scaling relations. The elementary power with \(p=2\) is called the square, \(\mathrm{pow}_2(x)=x^2\text{,}\) because it is the function that maps from the length of the side \(L\) to the area \(A\) of a square,

\begin{equation*} \mathrm{pow}_2 : L \mapsto A = L^2\text{.} \end{equation*}

The elementary power with \(p=3\) is called the cube, \(\mathrm{pow}_3(x)=x^3\text{,}\) because it is the function that maps from the length of the side \(L\) to the volume \(V\) of a cube,

\begin{equation*} \mathrm{pow}_3 : L \mapsto V = L^3\text{.} \end{equation*}

For other geometric shapes, the scaling relations involve an additional constant multiple. The area of a circle, for example, is a power function of radius with \(p=2\) but with a constant multiple \(\pi\text{,}\)

\begin{equation*} r \mapsto A = \pi r^2\text{.} \end{equation*}

The volume of a sphere is also a power function of radius with \(p=3\text{,}\)

\begin{equation*} r \mapsto V = \frac{4\pi}{3} r^3\text{.} \end{equation*}

Subsection 2.4.3 Inverse Functions

When we think of a function as a map between variables, say \(f : A \mapsto C\text{,}\) we think of \(f\) as the rule that goes from an input value on the \(A\) number line to a corresponding predicted output value on the \(C\) number line. An inverse function would be a rule that goes in the reverse direction, \(C \mapsto A\text{.}\) A function and its inverse function allow us to go back and forth between the two variables in either direction.

Definition 2.4.6

A function representing a map \(f : A \mapsto C\) has an inverse function, which we write \(f^{-1} : C \mapsto A\text{,}\) if the equation \(f(a)=c\) is equivalent to \(f^{-1}(c)=a\) for every state \((A,C)=(a,c)\text{.}\)  1 This condition on the state will correspond to a requirement on the domain and range of a function and its inverse.

We first illustrate the idea of an inverse function with a function defined by a simple map and no formula.

Example 2.4.7

Imagine a theater that has a promotional wheel so that the price of a ticket is based on which number you spin. The prices are listed in the table below.

Spin Price
1 $2
2 $5
3 $8
4 $10

We introduce variables for the system. Let \(S\) represent the result of the spin and let \(P\) represent the price of a ticket. The map \(f : S \mapsto P\) can be visualized using number lines.

The inverse map \(f^{-1} : P \mapsto S\) reverses the direction of the arrows.

Based on this system, we see that \(f(1)=2\) because when \(S=1\) we have \(P=2\text{.}\) The equivalent inverse equation is \(f^{-1}(2)=1\) because a price \(P=2\) comes from \(S=1\text{.}\) Similarly, \(f(2)=5\) and \(f^{-1}(5)=2\) are equivalent. Because the system is defined by the table and not a formula, \(f(5)\) and \(f^{-1}(3)\) each have no meaning. In the first case, \(f(5)\) has no meaning because \(S=5\) is not a possibile spin. In the second case, \(f^{-1}(3)\) has no meaning because \(P=3\) is not a possible ticket price.

You might have realized a possible problem. What happens if two input values map to the same output value? We wouldn't know which arrow to follow for the reverse mapping. A function that guarantees that different input values always have different output values is called one-to-one. A function that is not one-to-one has at least one value that is the output to two or more different input values.

When an algebraic equation defines the relation between the variables, we can attempt to solve the equation for either of the variables. If both variables can successfully be written as dependent variables, the corresponding formulas define the inverse functions.

Example 2.4.9

A rope of length 100 centimeters is cut into exactly five pieces Two of the pieces are of one length, and the other three pieces are of another length. Let \(d\) be the length of the ropes in the group of two. Let \(t\) be the length of the ropes in the group of three.

Find the functions \(f : d \mapsto t\) and \(g : t \mapsto d\text{.}\) Interpret the meaning of \(f(10)\) and \(g(10)\text{.}\)


We start by finding an equation relating \(d\) and \(t\text{.}\) The total length of the five pieces of rope added together must equal the original length of rope. This results in an equation

\begin{equation*} 2d + 3t = 100\text{.} \end{equation*}

With this equation, we can solve for each of the state variables in turn. Solving for \(d\text{,}\) we subtract \(3t\) and divide by \(2\text{:}\)

\begin{equation*} d = \frac{100-3t}{2}\text{.} \end{equation*}

Solving for \(t\text{,}\) we subtract \(2d\) and divide by \(3\text{:}\)

\begin{equation*} t = \frac{100-2d}{3}\text{.} \end{equation*}

The function \(f\) was defined as the map \(d \mapsto t\text{,}\) so we use the equation with \(t\) as the dependent variable,

\begin{equation*} t = f(d) = \frac{100-2d}{3}\text{.} \end{equation*}

We can find and interpret \(f(10)\text{.}\) With \(10\) as an input, we find \(f(10) = \frac{100-2(10)}{3} = \frac{80}{3} = 26 \frac{2}{3}\text{.}\) To interpret this, we recall that the input represents a value for \(d\text{.}\) The equation represents the state \(d=10\) and \(t=26 \frac{2}{3}\) If the group of two has length 10 centimeters, then the group of three has length \(26 \frac{2}{3}\) centimeters.

The function \(g\) was defined as the map \(t \mapsto d\text{,}\) so we now use the equation with \(d\) as the dependent variable,

\begin{equation*} d = g(t) = \frac{100-3t}{2}\text{.} \end{equation*}

With \(t=10\) as an input, we find \(g(10) = \frac{100-3(10)}{2} = \frac{70}{2} = 35\text{.}\) The function tells us that \(d=35\) when \(t=10\text{.}\) The group of two has length 35 centimeters whenever the group of three has length \(10\) centimeters.

The functions \(f\) and \(g\) are inverse functions to each other. If we used a placeholder variable instead of the state variables, we would write

\begin{equation*} f(x) = \frac{100-2x}{3} \end{equation*}

with its inverse function

\begin{equation*} f^{-1}(x) = g(x) = \frac{100-3x}{2}\text{.} \end{equation*}

Similarly, \(g^{-1}(x) = f(x)\text{.}\)

Subsection 2.4.4 Roots and Logarithms

Power and exponential functions have corresponding inverse functions. A root provides the inverse operation to an integer power. A logarithm provides the inverse operation to an exponential.

Definition 2.4.10

For an integer \(n \gt 1\text{,}\) the \(n\)th root \(y=\sqrt[n]{x}\) is the value such that \(y^n=x\text{.}\) If \(n\)is even, we require \(x \ge 0\) and \(y \ge 0\text{.}\) If \(n\) is odd, there is no restriction.

For even values \(n\) and \(x \gt 0\text{,}\) the equation \(y^n=x\) has two solutions, \(y = \pm \sqrt[n]{x}\text{,}\) because multiplying an even number of negative values is positive, \((-1)^n=1\text{.}\) There are no real solutions to \(y^n=x\) when \(x \lt 0\) for an even power \(n\text{.}\)

A root can be written as a fractional power. The properties of exponents imply that \((x^p)^n = x^{pn}\text{.}\) If \(p=\frac{1}{n}\text{,}\) then \((x^{\frac{1}{n}})^n = x\text{.}\) That is, \(x^{\frac{1}{n}} = \sqrt[n]{x}\text{.}\) This equivalence means that for any rational number \(p=\frac{k}{n}\text{,}\) the power \(x^p\) can be computed using integer powers (repeated multiplication) and extracting roots:

\begin{equation*} x^{\frac{k}{n}} = (\sqrt[n]{x})^k\text{.} \end{equation*}
Definition 2.4.11

For any base \(b\) with \(b \gt 0\) and \(b \ne 1\text{,}\) the elementary exponential \(\exp_b(x)=b^x\) has an inverse function, \(\log_b(x)\text{,}\) which is called the logarithm of base \(b\text{.}\) The value \(y=\log_b(x)\) is defined for \(x \gt 0\) as that value \(y\) such that \(b^y = x\text{.}\)

Notice that both roots and logarithms are defined through the equation that they solve. We will often convert equations using these inverses to simplify our work. Inverse functions can be used in balanced operations to create equivalent equations in order to remove the operations they undo from one side of the equation.

Example 2.4.12

Solve \(\sqrt[3]{x}=2\text{.}\)


The equation \(\sqrt[3]{x} = 2\) has an isolated cube root on the left. The inverse operation is cubing, so we get an equivalent equation by cubing both sides.

\begin{gather*} \sqrt[3]{x}=2\\ (\sqrt[3]{x})^3 = 2^3\\ x=2^3 \end{gather*}

The solution is \(x=8\text{.}\)

Example 2.4.13

Solve \(x^4=4\text{.}\)


The equation \(x^4 = 4\) has an isolated integer power on the left. The inverse operation is a fourth root. Because the power is even, there are two solutions.

\begin{gather*} x^4=4\\ \sqrt[4]{x^4} = \sqrt[4]{4}\\ x = \pm \sqrt[4]{4} \end{gather*}

Because \(4=2^2\text{,}\) we could rewrite this as

\begin{equation*} x = \pm(2^2)^{\frac{1}{4}} = \pm 2^{\frac{2}{4}} = \pm 2^{\frac{1}{2}} = \pm \sqrt{2}\text{.} \end{equation*}
Example 2.4.14

Solve \(\log_3(x)=2\text{.}\)


The equation \(\log_3(x)=2\) has an isolated logarithm. The inverse operation is an exponential with the same base \(b=3\text{.}\) An equivalent equation is formed by applying this exponential to both sides of the equation.

\begin{gather*} \log_3(x)=2\\ \exp_3(\log_3(x)) = \exp_3(2) = 3^2\\ x = 9 \end{gather*}

This is saying that the equation \(\log_3(9)=2\) is equivalent to \(3^2=9\text{.}\)

Example 2.4.15

Solve \(4^x=8\text{.}\)


The equation \(4^x=8\) has an isolated exponential. The inverse operation is a logarithm with the same base \(b=4\text{.}\)

\begin{gather*} 4^x=8\\ \log_4(\exp_4(x)) = \log_4(8)\\ x = \log_4(8) \end{gather*}

We will learn techniques for simplifying logarithms later. Most scientific calculators only have logarithms for base \(b=10\) and for base \(b=e\text{.}\) The logarithm for \(b=10\) is called the common logarithm and appears on a calculator with out a base log. The logarithm for \(b=e\) is called the natural logarithm and appears on a calculator as ln. We will later prove that every logarithm can be found using one of these by the change of base formula

\begin{equation*} \log_b(x) = \frac{\log(x)}{\log(b)} = \frac{\ln(x)}{\ln(b)}\text{.} \end{equation*}

Subsection 2.4.5 Inverse Questions for Sequences

In our discussions about sequences, we introduced two maps. The first was in relation to an explicit definition of a sequence, where we mapped from the index to the sequence value, \(n \mapsto x_n\text{.}\) The second was in relation to a recursive definition of a sequence by a projection function, \(x_{n-1} \mapsto x_{n}\text{.}\) The inverse map for these functions can help answer interesting questions for sequences.

Example 2.4.16

Suppose you start with one cent. Each day your money doubles. When will you have a million dollars?


We start by identifying our sequence. We might choose the name \(W\) for our sequence to represent our wealth. Because the sequence doubles each day, our index \(n\) will represent the day and the sequence has recursive equation \(W_{n} = 2 W_{n-1}\text{.}\) This means our sequence is a geometric sequence. The initial value is \(W_1 = 0.01\) (dollars). We can then write down the explicit formula using Theorem 2.2.10,

\begin{equation*} W_n = 0.01 \cdot 2^{(n-1)}\text{.} \end{equation*}

The properties of exponents allow us to rewrite this as an equivalent expression

\begin{equation*} W_n = 0.01 \cdot \frac{2^n}{2^1} = 0.005 \cdot 2^n\text{.} \end{equation*}

The question is when the value of our wealth is a million dollars. That is, we want to know the day \(n\) when \(W_n = 1000000\text{.}\) This is an inverse function question. To answer a single question like this, we do not need to know the actual function. We can just solve for \(n\text{,}\) using balanced operations until \(n\) is isolated.

\begin{gather*} W_n = 0.005 \cdot 2^n\\ 0.005 \cdot 2^n = 10^6\\ 2^n=\frac{10^6}{0.005}\\ 2^n = 2 \times 10^8 \end{gather*}

On the left, we have an elementary exponential with base \(b=2\text{.}\) To isolate \(n\text{,}\) we need to use the logarithm with base \(b=2\text{.}\)

\begin{gather*} \log_2( 2^n) = \log_2( 2 \times 10^8 ) \\ n = \log_2( 2 \times 10^8 ) \end{gather*}

Using the change of base rule for logarithms, we can use a calculator to identify the value.

\begin{equation*} n = \frac{\log(2 \times 10^8)}{\log(2)} \approx 27.5754 \end{equation*}

We try to interpret the result. The value \(n \approx 27.5754\) is not possible an index, which must be an integer. That means that our wealth is never exactly a million dollars. We need to reinterpret the question as saying when we first exceed a million dollars. Because our solution is between \(n=27\) and \(n=28\) and the value doubles each day, we can see that day \(n=27\) will be the last day we have a wealth below a million dollars and that day \(n=28\) will be the first day when our wealth is above a million dollars.

Example 2.4.17

A population of insects is tracked weekly and found to follow a recursive model. We let \(P\) be the sequence of population measurements with an index \(n\) counting the number of weeks. The insect population follows the recurrence relation

\begin{equation*} P_{n} = 1.5 P_{n-1} - 100 \end{equation*}

with a value of \(P_0 = 5000\text{,}\) the population recorded for the first week of observation. What was the population two weeks before observations began?


The recurrence relation shows that \(P\) is not arithmetic nor is it geometric. We do not yet know how to find an explicit formula. However, we do know the projection function \(f : P_{n-1} \mapsto P_{n}\text{.}\) The inverse of the projection function would reverse the direction of projection, allowing us to determine the preceding value of the sequence instead of the subsequent value.

We find the inverse function (the reverse projection function) by solving the recurrent relation for \(P_n\text{.}\)

\begin{gather*} P_{n} = 1.5 P_{n-1} - 100\\ P_{n} + 100 = 1.5 P_{n-1} = \frac{3}{2} P_{n-1}\\ \frac{2}{3}(P_{n} + 100) = P_{n-1} \end{gather*}

Consequently, the inverse projection function is \(f^{-1}(x) = \frac{2}{3}(x+100)\text{.}\)

With the inverse projection function, we can work the sequence backwards from the initial value. Starting with \(P_0 = 5000\text{,}\) we find

\begin{gather*} P_{-1} = f^{-1}(5000) = \frac{2}{3}(5000+100) = 3400\\ P_{-2} = f^{-1}(3400) = \frac{2}{3}(3400+100) = \frac{7000}{3} \end{gather*}

The exact value from our model is \(P_{-2} \approx 2333\frac{1}{3}\text{.}\) Even though a population has integer values and this prediction is definitely not an integer, we would report our model value unless we had information on how many significant digits to keep.

Subsection 2.4.6 Summary

  • Properties of exponents are motivated by the idea that integer powers correspond to repeated multiplication.
  • Exponential functions (exponential growth or exponential decay) have the form \(f(x) = A \cdot b^x\text{,}\) with the independent variable in the exponent.
  • Power functions have the form \(f(x) = A \cdot x^p\text{,}\) with the independent variable as the base of the power.
  • The inverse for a function \(f : A \mapsto C\) is a function \(f^{-1} : C \mapsto A\text{.}\) This requires that \(f\) is one-to-one, so that different inputs always result in different outputs.
  • We use inverse functions to solve equations by creating equivalent equations that eliminate a function from one side:
    \begin{equation*} f(x) = y \qquad \Leftrightarrow \qquad x = f^{-1}(y). \end{equation*}
  • Roots, such as the square root or cube root, are inverse functions for elementary power functions:
    \begin{equation*} x^n = y \qquad \Leftrightarrow \qquad x = \sqrt[n]{y}\text{.} \end{equation*}
    When \(n\) is an even integer, we require \(x \ge 0\) and \(y \ge 0\text{.}\)
  • Roots can be written as reciprocal powers:
    \begin{equation*} \sqrt[n]{x} = x^{1/n}\text{.} \end{equation*}
  • Logarithms are inverse functions for elementary exponential functions, defined for every base \(b \gt 0\) and \(b \ne 1\text{:}\)
    \begin{equation*} b^x = y \qquad \Leftrightarrow \qquad x = \log_b(y)\text{.} \end{equation*}
  • The change of base rule allows us to find decimal approximations for any base using the common or natural logarithm:
    \begin{equation*} \log_b(x) = \frac{\log(x)}{\log(b)} = \frac{\ln(x)}{\ln(b)}\text{.} \end{equation*}

Subsection 2.4.7 Exercises

Identify a property of exponents to rewrite an equivalent expression. Note that each property can be applied in either direction.






\(2^x \cdot 3^x\)


\(\displaystyle \frac{x^3}{4^3}\)


\(\displaystyle \frac{3^u}{3^4}\)


\(\displaystyle 2^{3x}\)


A student made a mistake writing \(\displaystyle 3 \cdot 2^x = 6^x\text{.}\) What did the student do? Why was it incorrect?

In each problem, a relation between two variables is defined by a table of data. Illustrate the relations using arrows between two number lines. In which direction does the relation define a map corresponding to a function? Does that function have an inverse function? Explain why or why not.

\(A\) \(C\)
0 5
1 8
2 9
3 8
4 11
\(x\) \(E\)
0 1
1 2
2 4
3 8
4 16
\(t\) \(Q\)
0 8
1 5
2 2
0 -1
1 -4
2 -7

For each equation defining a relation between two variables, find the pair of inverse functions. Identify each function as a map, showing clearly the input and output variables.






\(2rs - 6 = 4r - 3s\)


\(P=20 \cdot 2^{t/8}\)

Solve the equations.


\(x^7 = 4\)




\(\sqrt[4]{x} = 3\)


\(3 \sqrt[3]{2x} = 4\)










\(4 \log_5(x)= 15\)


\(3x^2 - x^6 = 0\)


\(4 \cdot x^5 = 3\)


\(4 \cdot 5^x = 3\text{,}\) writing solutions in terms of the natural logarithm ln.


\(2 \cdot 3^{-x} = 3 \cdot 2^{x}\text{,}\) writing solutions in terms of the natural logarithm ln. Hint: Find an equivalent equation with a single exponential after using properties of exponents.



Temperature can be measured in degrees Celsius and degrees Fahrenheit. Let \(C\) be the temperature measured in degrees Celsius, and let \(F\) be the temperature measured in degrees Fahrenheit. Under standard conditions, water freezes at the state \((C,F) = (0,32)\) and boils at the state \((C,F)=(100,212)\text{.}\) The relation between \(F\) and \(C\) is a linear relation.

Answer the following questions.

  1. Explain what the states \((C,F)=(0,32)\) and \((C,F)=(100,212)\) means in terms of individual variables.
  2. Show that both given states satisfy the equation \(5F-9C=160\text{.}\)
  3. Find explicit functions \(g : F \mapsto C\) and \(h : C \mapsto F\text{.}\)
  4. Compute \(g(45)\text{.}\) What does this calculation represent?
  5. Compute \(h(45)\text{.}\) What does this calculation represent?