Exponentials and Logarithms — AP® Calculus AB Notes

Some quantities grow or shrink by the same percentage in each equal stretch of time: a savings account, a bacteria culture, a drug leaving the blood, the gap between a cup of coffee's temperature and the room's. These are modeled by exponential functions, and calculus has a favorite one, $e^x$ . Its partner, the natural logarithm $\ln x$ , lets you solve for a time hidden in an exponent. This page reviews both and the rules that connect them.

Exponential functions and the number $e$

An exponential function has the variable in the exponent: $f(x) = b^x$ , with a positive base $b \ne 1$ . Each step of 1 in $x$ multiplies the output by $b$ . With $b > 1$ the function shows exponential growth; with $0 < b < 1$ , exponential decay.

Calculus works best with one particular base, $e \approx 2.71828$ . The reason will appear in Unit 2: $e^x$ is the function whose rate of change equals its own value, which makes every formula involving it clean. For now, know its graph and its rules:

$e^0 = 1$ , so the graph passes through $(0, 1)$ .
$e^x > 0$ for every $x$ . The graph never touches the $x$ -axis; the $x$ -axis is a horizontal asymptote on the left.
All the exponent rules apply: $e^a e^b = e^{a + b}$ , $\frac{e^a}{e^b} = e^{a - b}$ , $(e^a)^b = e^{ab}$ , $e^{-x} = \frac{1}{e^x}$ .

The natural logarithm

The natural logarithm $\ln x$ answers one question: what power of $e$ gives $x$ ? $\ln x = y \quad \text{means} \quad e^y = x$ So $\ln e = 1$ , $\ln 1 = 0$ , and $\ln(e^5) = 5$ . Since $e^y$ is always positive, $\ln x$ is defined only for $x > 0$ .

Because $\ln$ undoes $e^x$ and $e^x$ undoes $\ln$ , they cancel each other: $e^{\ln x} = x \;(x > 0) \qquad \ln(e^x) = x \;(\text{every } x)$

The curve y = e to the x passes through (0, 1) and (1, e) and stays above the x-axis; the curve y = ln x passes through (1, 0) and (e, 1) and exists only for x greater than 0. They are mirror images across the dashed line y = x. — **Figure 1.** Swap the coordinates of every point on $y = e^x$ and you get $y = \ln x$ : a reflection across the line $y = x$ . LevlPrep original diagram.

Log rules

Every log rule is an exponent rule read backward. Multiplying powers adds exponents, so the log of a product is a sum:

Rule	Example
$\ln(ab) = \ln a + \ln b$	$\ln 12 = \ln 3 + \ln 4$
$\ln\frac{a}{b} = \ln a - \ln b$	$\ln\frac{x^3}{y} = 3\ln x - \ln y$
$\ln(a^p) = p\ln a$	$\ln\sqrt{x} = \frac{1}{2}\ln x$

And the rules that do not exist, which are the classic slips:

Wrong	Why
$\ln(a + b) = \ln a + \ln b$	a sum of logs comes from a product: $\ln a + \ln b = \ln(ab)$
$(\ln x)^2 = 2\ln x$	$2\ln x = \ln(x^2)$ , a different thing
$\frac{\ln a}{\ln b}$ $= \ln(a - b)$	there is no rule for a quotient of logs

Worked example 1. Simplify $\ln\left(x^2\sqrt{x}\right)$ and $e^{2\ln 3}$ .

$x^2\sqrt{x}$ $= x^2 \cdot x^{1/2}$ $= x^{5/2}$ , so $\ln\left(x^{5/2}\right) = \frac{5}{2}\ln x$ .

$2\ln 3 = \ln 3^2 = \ln 9$ , so $e^{2\ln 3} = e^{\ln 9} = 9$ . Move the coefficient into the log before canceling $e$ and $\ln$ .

Solving exponential and log equations

To free a variable from an exponent, isolate the exponential, then take $\ln$ of both sides. To free a variable from inside a log, isolate the log, then raise $e$ to both sides.

Worked example 2. Solve $3e^{2x} = 21$ .

Isolate: $e^{2x} = 7$ . Take $\ln$ : $2x = \ln 7$ . Solve: $x = \frac{\ln 7}{2}$ .

A common slip is dividing by 2 inside the log: $\ln\frac{7}{2}$ is a different number.

Worked example 3. Solve $\ln(x - 1) = 3$ .

Raise $e$ to both sides: $x - 1 = e^3$ , so $x = e^3 + 1$ .

Check the domain: $x - 1 = e^3 > 0$ , so the solution is valid.

Growth and decay with $e$

A quantity that starts at $A$ and grows or decays continuously at a steady percentage rate is written $A e^{kt}$ . The number $k$ sets the speed: $k > 0$ for growth, $k < 0$ for decay. Two questions come up constantly: when does it reach a level, and how long does it take to double or halve?

Worked example 4. A sample of $A(t) = 80e^{kt}$ grams halves every 5 hours. Find $k$ .

$80e^{5k} = 40$ , so $e^{5k} = \frac{1}{2}$ , so $5k = \ln\frac{1}{2} = -\ln 2$ .

$k = -\frac{\ln 2}{5} \approx -0.139$ per hour. The negative sign confirms decay.

On a calculator question, store $\ln 2$ and other intermediate values rather than rounding them, and give the final answer to three decimal places.

Exponential functions and the number ee

The natural logarithm

Log rules

Solving exponential and log equations

Growth and decay with ee

Exponential functions and the number $e$

Growth and decay with $e$