Topic 2: Exponential and Logarithmic Functions

Unit 2 of AP Precalculus shifts your attention from the additive, polynomial world of Unit 1 to the multiplicative world of exponential and logarithmic functions. The central idea is constant proportional change: where a linear function adds a fixed amount over equal input intervals, an exponential function multiplies by a fixed factor. Sequences are the discrete gateway to this idea, with arithmetic sequences mirroring linear growth and geometric sequences mirroring exponential growth. Once you understand that logarithms are simply the inverse operation that recovers an exponent, the entire unit becomes a connected story about modeling real phenomena, manipulating expressions with properties, solving equations, and revealing hidden structure with semi-log plots. This article walks through each strand and shows how they fit together for the exam.

Arithmetic and Geometric Sequences

A sequence is a function whose domain is the positive integers (or sometimes the non-negative integers). An arithmetic sequence has a constant difference between consecutive terms: each term is found by adding a fixed common difference $d$. Its explicit form is $a_n = a_1 + (n-1)d$ and its recursive form is $a_n = a_{n-1} + d$ with a given first term. Because a fixed amount is added per step, arithmetic sequences behave like linear functions, and their graphs are evenly spaced collinear points. A geometric sequence has a constant ratio between consecutive terms: each term is the previous term multiplied by a fixed common ratio $r$. Its explicit form is $g_n = g_1 \cdot r^{n-1}$ and its recursive form is $g_n = r \cdot g_{n-1}$. Because a fixed factor is multiplied per step, geometric sequences behave like exponential functions. When $\lvert r \rvert \gt 1$ the terms grow in magnitude; when $0 \lt \lvert r \rvert \lt 1$ they decay toward zero. A frequent exam subtlety is indexing. Notice the exponent is $n-1$, not $n$, when the first term carries index $1$. If a problem instead starts at $n = 0$, the formula shifts to $g_n = g_0 \cdot r^{n}$. Always confirm which term corresponds to which index before plugging in.

Exponential Functions and Constant Proportional Change

An exponential function has the form $f(x) = a \cdot b^{x}$, where $a$ is the initial value (the output at $x = 0$) and $b \gt 0$, $b \neq 1$, is the base or growth factor. The defining property is constant proportional change: over equal-length input intervals the output is multiplied by the same factor. Equivalently, the ratio $\frac{f(x_2)}{f(x_1)}$ depends only on the difference $x_2 - x_1$, not on the starting point. Contrast this with linear functions, where equal input intervals produce equal differences in output. A simple test on a data table distinguishes them: if successive outputs share a common difference, the model is linear; if successive outputs share a common ratio, the model is exponential. The natural exponential function uses base $e$, where $e \approx 2.718$. The form $f(x) = a \cdot e^{kx}$ is common in continuous growth models. Here $e^{k}$ plays the role of $b$, so a positive $k$ gives growth and a negative $k$ gives decay. Exponential functions are always positive when $a \gt 0$, have a horizontal asymptote at $y = 0$ (before any vertical shift), and are one-to-one, which is exactly what guarantees they have inverses.

Growth, Decay, and Transformations

When $b \gt 1$ the function models growth; when $0 \lt b \lt 1$ it models decay. A useful reparametrization writes growth as $f(t) = a(1 + r)^{t}$ and decay as $f(t) = a(1 - r)^{t}$, where $r$ is the constant percent rate per period expressed as a decimal. For example, a population growing 8 percent per year is modeled by $a(1.08)^{t}$, while a substance losing 8 percent per year is modeled by $a(0.92)^{t}$. Transformations follow the general rules. The function $f(x) = a \cdot b^{(x-h)} + k$ shifts the parent graph horizontally by $h$ and vertically by $k$. The vertical shift $k$ moves the horizontal asymptote to $y = k$, which is a detail students often miss. A negative leading coefficient reflects the graph across the horizontal asymptote. An important conceptual point for the exam: a horizontal shift of an exponential function is equivalent to a vertical dilation, because $b^{(x-h)} = b^{-h} \cdot b^{x}$. The factor $b^{-h}$ is just a constant multiplier. This is unique to exponential functions and reflects their multiplicative nature.

Logarithms and Their Properties

A logarithm answers the question: to what exponent must the base be raised to obtain a given number? Formally, $\log_b(y) = x$ is equivalent to $b^{x} = y$, for $b \gt 0$, $b \neq 1$, and $y \gt 0$. The common logarithm $\log$ uses base $10$, and the natural logarithm $\ln$ uses base $e$. The key properties, which all follow from exponent rules, are essential and frequently tested. The product property is $\log_b(xy) = \log_b x + \log_b y$. The quotient property is $\log_b\left(\frac{x}{y}\right) = \log_b x - \log_b y$. The power property is $\log_b(x^{n}) = n \log_b x$. Two identities follow directly from the definition: $\log_b(b^{x}) = x$ and $b^{\log_b x} = x$, capturing the inverse relationship. The change-of-base formula lets you evaluate any logarithm with a calculator: $\log_b x = \frac{\log x}{\log b} = \frac{\ln x}{\ln b}$. A common error is to misapply properties: $\log_b(x + y)$ does NOT equal $\log_b x + \log_b y$, and $\frac{\log_b x}{\log_b y}$ does NOT equal $\log_b\left(\frac{x}{y}\right)$. Memorize the legitimate forms and resist the temptation to invent others.

Logarithmic Functions, Graphs, and Inverses

Because exponential functions are one-to-one, they have inverse functions, and those inverses are the logarithmic functions. If $f(x) = b^{x}$ then $f^{-1}(x) = \log_b x$. Inverse functions undo each other, so $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$ wherever the compositions are defined. Graphically, a function and its inverse are reflections of each other across the line $y = x$. This reflection swaps the features of the two graphs. The exponential function has domain all real numbers, range $y \gt 0$, and a horizontal asymptote at $y = 0$. The logarithmic function therefore has domain $x \gt 0$, range all real numbers, and a vertical asymptote at $x = 0$. The exponential passes through $(0, 1)$; the logarithm passes through $(1, 0)$. A logarithmic function $g(x) = \log_b x$ with $b \gt 1$ is increasing but increasingly slowly, illustrating that logarithmic growth is the slowest of the common models. Transformations apply as usual: $g(x) = a \log_b(x - h) + k$ shifts the vertical asymptote to $x = h$. Always check the domain after a horizontal shift, since the argument of a logarithm must stay positive.

Solving Exponential and Logarithmic Equations

To solve an exponential equation, isolate the exponential expression and then take a logarithm of both sides. For example, to solve $5 \cdot 2^{x} = 80$, divide to get $2^{x} = 16$, then apply a logarithm: $x = \log_2 16 = 4$. When the answer is not a clean power, use the power property: from $3^{x} = 20$ you get $x \ln 3 = \ln 20$, so $x = \frac{\ln 20}{\ln 3}$. To solve a logarithmic equation, consolidate logarithms into a single logarithm using the properties, then rewrite in exponential form. For instance, $\log_2(x) + \log_2(x - 2) = 3$ becomes $\log_2\left(x(x-2)\right) = 3$, hence $x(x-2) = 2^{3} = 8$, giving $x^{2} - 2x - 8 = 0$ and $x = 4$ or $x = -2$. The critical final step is checking for extraneous solutions. Logarithms require positive arguments, so $x = -2$ must be rejected because $\log_2(-2)$ is undefined. Only $x = 4$ is valid. Whenever you solve a logarithmic equation, substitute each candidate back to confirm every logarithm argument is positive.

Comparing Models, Composition, and Semi-Log Plots

A recurring theme is comparing function behavior. For large inputs, an increasing exponential function eventually outgrows any polynomial function, no matter how large the polynomial degree, and a logarithmic function eventually grows slower than any positive-power polynomial. Recognizing the right model from a context or data set is a core skill: look for constant differences (linear), constant second differences (quadratic), constant ratios (exponential), or rapid early change that flattens (logarithmic). Composition combines functions, and the exponential and logarithmic pair are inverses, so composing them cancels: $\ln(e^{x}) = x$ and $e^{\ln x} = x$. Composition also lets you rewrite models, for example expressing any exponential base in terms of $e$ via $b^{x} = e^{(\ln b) x}$. Semi-log plots are a powerful linearization tool. If you plot the logarithm of the output against the input, an exponential model $y = a b^{x}$ becomes linear, because taking logs gives $\log y = \log a + (\log b) x$. This is a line with slope $\log b$ and intercept $\log a$. So data that looks straight on a semi-log plot signals exponential behavior, and you can recover $a$ and $b$ from the intercept and slope. By contrast, a power model $y = a x^{n}$ becomes linear on a log-log plot, where both axes are logarithmic.