Logarithmic Functions

AP Precalculus· AP Precalculus CED — Exponential and Logarithmic Functions· 14 min read

1. Definition of Logarithmic Functions as Inverses★★☆☆☆⏱ 3 min

A logarithmic function is the inverse of a one-to-one exponential function. Formally, for $b > 0, b \neq = 1$ , $y = lo g_{b} x$ if and only if $b^{y} = x$ . Standard notation uses $lo g x$ for the common logarithm (base 10) and $ln x$ for the natural logarithm (base $e \approx 2.71828$ ), the most common form for calculus and continuous growth models.

Logarithmic functions reverse exponential operations, allowing us to solve for unknown exponents and model logarithmic scaling phenomena, from pH levels to decibel ratings. As the inverse of exponential functions, logarithmic functions swap the domain and range of exponential functions: logarithmic functions have a domain of only positive real numbers ( $x > 0$ ) and a range of all real numbers.

📘 Definition

Logarithmic Function

$y = lo g_{b} x$ , $ln x$ (base $e$ ), $lo g x$ (base 10)

The inverse of a one-to-one exponential function with base $b > 0, b \neq = 1$ , satisfying $lo g_{b} x = y ⟺ b^{y} = x$ , defined only for positive $x$ .

Example:

$lo g_{2} 8 = 3$ , since $2^{3} = 8$

📐 Worked Example

Rewrite $3^{4} = 81$ in logarithmic form, and rewrite $lo g_{4} 64 = 3$ in exponential form.

1
The base of the exponential becomes the base of the logarithm, and the exponent becomes the output of the logarithm.
$3^4 = 81 \implies \log_3 81 = 4$
2
For the logarithmic statement, the base of the logarithm becomes the base of the exponential, and the output becomes the exponent.
$\log_4 64 = 3 \implies 4^3 = 64$

2. Core Properties and Change of Base Formula★★☆☆☆⏱ 4 min

All logarithm properties are derived directly from corresponding exponent rules, due to the inverse relationship between logarithms and exponentials. The key inverse identities connecting the two function types are:

b^{\log_b x} = x \quad (x>0) \quad \text{and} \quad \log_b (b^x) = x \quad (\text{for all real } x)

Product Rule: $lo g_{b} (M N) = lo g_{b} M + lo g_{b} N$ (turns products of positive numbers into sums of logarithms)
Quotient Rule: $lo g_{b} (\frac{M}{N}) = lo g_{b} M - lo g_{b} N$ (turns quotients of positive numbers into differences of logarithms)
Power Rule: $lo g_{b} (M^{k}) = k lo g_{b} M$ for any real $k$ and positive $M$

The change of base formula allows you to evaluate any logarithm using a standard calculator, which only computes base 10 or base $e$ logarithms:

\log_b a = \frac{\ln a}{\ln b} = \frac{\log_{10} a}{\log_{10} b} \quad (a>0, b>0, b \neq 1)

📐 Worked Example

Simplify $lo g_{5} (250) - lo g_{5} (2) + 3 lo g_{5} (5)$ to a simplified numerical value.

1
Apply the quotient rule to the first two terms:
$\log_5 \left(\frac{250}{2}\right) + 3 \log_5 (\sqrt{5}) = \log_5 (125) + 3 \log_5 (5^{1/2})$
2
Apply the power rule to the second term, using $lo g_{b} b = 1$ :
$3 \log_5 (5^{1/2}) = 3 \cdot \frac{1}{2} \cdot \log_5 5 = \frac{3}{2} (1) = \frac{3}{2}$
3
Rewrite 125 as a power of 5 and apply the inverse identity:
$\log_5 (5^3) = 3 \log_5 5 = 3(1) = 3$
4
Add the results for the final value:
$3 + \frac{3}{2} = \frac{9}{2} = 4.5$

3. Solving Logarithmic Equations★★★☆☆⏱ 4 min

A logarithmic equation has an unknown variable inside the argument of a logarithm. The core solution strategy uses the inverse relationship between logs and exponentials: isolate a single logarithmic term, convert to exponential form, then solve for the variable. If you have multiple logarithms on the same side, combine them using logarithm properties first.

The most critical step often missed is checking for extraneous solutions. Combining multiple logarithms erases individual domain restrictions, so all candidate solutions must be checked against the original equation, and any solution that makes a logarithm's argument non-positive must be discarded.

📐 Worked Example

Solve $lo g_{2} (x + 2) + lo g_{2} (x) = 3$ for all real solutions.

1
Write domain restrictions from the original equation: $x + 2 > 0 ⟹ x > - 2$ and $x > 0$ , so overall valid domain is $x > 0$ .
2
Combine the two logarithms with the product rule:
$\log_2 \left(x(x+2)\right) = 3 \implies \log_2 (x^2 + 2x) = 3$
3
Convert to exponential form by definition of logarithms:
$2^3 = x^2 + 2x \implies x^2 + 2x - 8 = 0$
4
Factor the quadratic to get candidate solutions:
$(x+4)(x-2) = 0 \implies x=-4 \text{ and } x=2$
5
Check against the original domain: $x = - 4 < 0$ is extraneous and discarded, while $x = 2 > 0$ is valid.

✓ Quick check

Test your understanding of logarithm properties:

Which of the following is equivalent to $3 ln 2 - ln 4$ ?
- A) $ln 2$
- B) $ln 6$
- C) $2 ln 3$
- D) $- ln 2$
Reveal answer
A) $\ln 2$ —
Apply the power rule: $3 ln 2 = ln 2^{3} = ln 8$ , then apply the quotient rule: $ln 8 - ln 4 = ln (8/4) = ln 2$ .

4. Graphing and Transformations of Logarithmic Functions★★★☆☆⏱ 3 min

The parent logarithmic function $y = lo g_{b} x$ has consistent core features: it has a vertical asymptote at $x = 0$ (the y-axis), passes through $(1, 0)$ (since $lo g_{b} 1 = 0$ for any valid $b$ ), has domain $(0, \infty)$ and range of all real numbers. If $b > 1$ , the function is strictly increasing and concave down; if $0 < b < 1$ , it is strictly decreasing and concave up.

Transformations of logarithmic functions follow the same rules as all other function transformations: for $y = a lo g_{b} (c (x - h)) + k$ , $a$ controls vertical stretch/compression/reflection, $c$ controls horizontal stretch/compression/reflection, $h$ is the horizontal shift, and $k$ is the vertical shift. To find the new vertical asymptote after shifting, simply set the argument of the logarithm equal to zero and solve for $x$ , which avoids mistakes from memorizing shift directions.

📐 Worked Example

For $f (x) = ln (x - 3) + 2$ , identify the domain, equation of the vertical asymptote, and one point on the graph, then describe the transformation from the parent $y = ln x$ .

1
Find the domain by requiring the argument to be positive:
$x-3>0 \implies x>3, \text{ so domain is } (3, \infty)$
2
Find the vertical asymptote by setting the argument equal to zero:
$x-3=0 \implies x=3$
3
Find a simple point using the parent function's known intercept: when $ln (argument) = 0$ , the argument equals 1. Set $x - 3 = 1 ⟹ x = 4$ , then calculate $f (4)$ :
$f(4) = \ln(1) + 2 = 0 + 2 = 2, \text{ so } (4,2) \text{ is on the graph}$
4
Describe the transformation:
$\text{The graph of } y = \ln x \text{ is shifted 3 units right and 2 units up}$

📐 Worked Example

The pH of an aqueous solution is defined as $pH = - lo g_{10} [H^{+}]$ , where $[H^{+}]$ is hydrogen ion concentration in moles per liter (mol/L). A sample of orange juice has a pH of 3.8, and black coffee has a pH of 5.0. (a) Find the hydrogen ion concentration of orange juice (rounded to 3 significant figures). (b) How many times greater is the hydrogen ion concentration in orange juice than in black coffee?

1
Part (a): Substitute pH = 3.8 and rearrange:
$3.8 = -\log_{10} [H^+] \implies \log_{10} [H^+] = -3.8$
2
Rewrite in exponential form and calculate:
$[H^+] = 10^{-3.8} \approx 1.58 \times 10^{-4} \text{ mol/L}$
3
Part (b): Find the hydrogen ion concentration of black coffee:
$5.0 = -\log_{10} [H^+]_{\text{coffee}} \implies [H^+]_{\text{coffee}} = 10^{-5.0} = 1.0 \times 10^{-5} \text{ mol/L}$
4
Calculate the ratio of concentrations:
$\frac{1.58 \times 10^{-4}}{1.0 \times 10^{-5}} = 15.8 \approx 16$
5
Final result: The hydrogen ion concentration of orange juice is approximately 16 times greater than that of black coffee.

5. Common Pitfalls

Wrong move:

$lo g_{b} (M + N) = lo g_{b} M + lo g_{b} N$

Why:

Students confuse the product rule with adding arguments, misremembering that logs turn products into sums, not sums into sums.

Correct move:

Only $lo g_{b} (M N) = lo g_{b} M + lo g_{b} N$ ; there is no general simplification for the logarithm of a sum.

Wrong move:

After solving $lo g_{2} (x + 2) + lo g_{2} x = 3$ , keeping $x = - 4$ as a valid solution

Why:

Students check solutions against the final combined equation, not the original equation, forgetting that individual terms have domain restrictions lost when combining.

Correct move:

Always list domain restrictions from the original equation before solving, and discard any candidate that violates any restriction.

Wrong move:

$\frac{l o g _{b} M}{l o g _{b} N} = lo g_{b} (\frac{M}{N})$

Why:

Students confuse the change of base formula with the quotient rule, mixing up the order of operations.

Correct move:

The quotient rule is $lo g_{b} (\frac{M}{N}) = lo g_{b} M - lo g_{b} N$ , while $\frac{l o g _{b} M}{l o g _{b} N} = lo g_{N} M$ by change of base—they are not equivalent.

Wrong move:

Stating the domain of $f (x) = lo g (x^{2})$ is $x > 0$

Why:

Students generalize the domain rule for $lo g (x)$ to $lo g (x^{2})$ without checking when the argument is positive.

Correct move:

Always check for positive arguments directly: $x^{2} > 0$ for all $x \neq = 0$ , so domain is $(- \infty, 0) \cup (0, \infty)$ .

Wrong move:

Writing the vertical asymptote of $y = ln (x + 2) - 1$ as $x = 2$

Why:

Students mix up horizontal shift direction, assuming $+ 2$ shifts right instead of left.

Correct move:

Always find the vertical asymptote by setting the argument equal to zero and solving: $x + 2 = 0 ⟹ x = - 2$ , which gives the correct asymptote regardless of shift direction.

6. Quick Reference Cheatsheet

Category	Formula	Notes
Definition of Logarithm	$lo g_{b} y = x ⟺ b^{x} = y$	Requires $b > 0, b \neq = 1, y > 0$
Inverse Identity 1	$b^{l o g_{b} x} = x$	Only valid for $x > 0$ , any valid base $b$
Inverse Identity 2	$lo g_{b} (b^{x}) = x$	Valid for all real $x$ , any valid base $b$
Product Rule	$lo g_{b} (M N) = lo g_{b} M + lo g_{b} N$	Requires $M > 0, N > 0$ ; does not apply to $lo g_{b} (M + N)$
Quotient Rule	$lo g_{b} (\frac{M}{N}) = lo g_{b} M - lo g_{b} N$	Requires $M > 0, N > 0$ ; not equivalent to $\frac{l o g _{b} M}{l o g _{b} N}$
Power Rule	$lo g_{b} (M^{k}) = k lo g_{b} M$	Requires $M > 0$ , works for any real $k$
Change of Base Formula	$lo g_{b} a = \frac{l n a}{l n b} = \frac{l o g _{10} a}{l o g _{10} b}$	Used to evaluate any log with a calculator; $a > 0, b > 0, b \neq = 1$
Parent Log Graph Features	$y = lo g_{b} x$	Vertical asymptote at $x = 0$ , passes through $(1, 0)$ , domain $(0, \infty)$ , range $(- \infty, \infty)$ , increasing if $b > 1$ , decreasing if $0 < b < 1$

What's Next

Mastery of logarithmic functions is required for all remaining topics in Unit 2 of AP Precalculus, starting with exponential and logarithmic modeling, where you will use logarithms to solve for unknown parameters in continuous growth and decay models. Without solid proficiency in simplifying logarithmic expressions and checking for extraneous solutions when solving logarithmic equations, you will struggle to calculate half-life, doubling time, or estimate parameters for exponential data, a frequent free-response question topic on the AP exam. Logarithmic functions also lay critical groundwork for future calculus study, where the natural logarithm is the antiderivative of $\frac{1}{x}$ , and logarithmic differentiation simplifies complex derivative problems.