4.3 Logarithmic Functions

Logarithmic functions are the inverse of exponential functions, giving you a way to "undo" exponentiation. They show up constantly in modeling growth, decay, and scale across physics, chemistry, and finance, so getting comfortable with them now pays off throughout the course.

This section covers converting between logarithmic and exponential forms, evaluating logarithmic expressions, applying key properties, graphing techniques, and real-world applications like earthquake measurement and sound intensity.

Logarithmic Functions

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Logarithmic and exponential conversions

The core idea: a logarithm answers the question "what exponent do I need?" The expression $\log_b(x) = y$ is just another way of writing $b^y = x$. In both forms, $b$ is the base, $x$ is the argument (the result of the exponentiation), and $y$ is the exponent (the logarithm's output).

Because logarithmic and exponential functions are inverses of each other, they cancel out when composed:

• Exponential undoes log: $b^{\log_b(x)} = x$
  • Example: $2^{\log_2(8)} = 8$ because $\log_2(8) = 3$ and $2^3 = 8$
• Log undoes exponential: $\log_b(b^x) = x$
  • Example: $\log_5(5^2) = 2$ because $5^2 = 25$ and $\log_5(25) = 2$

When you're stuck on a log problem, try rewriting it in exponential form (or vice versa). That single move solves most conversion questions.

Evaluation of logarithmic expressions

Common logarithm $\log_{10}(x)$ is written simply as $\log(x)$ (no base shown means base 10).

• $\log(1000) = 3$ because $10^3 = 1000$
• $\log(0.01) = -2$ because $10^{-2} = 0.01$

Natural logarithm $\log_e(x)$ is written as $\ln(x)$. The base $e \approx 2.71828$ is a mathematical constant that appears naturally in continuous growth and calculus.

• $\ln(e^4) = 4$ (log undoes the exponential)
• $\ln(1) = 0$ because $e^0 = 1$

Change of base formula lets you evaluate any logarithm using a base your calculator supports (base 10 or base $e$):

$\log_b(x) = \frac{\log_a(x)}{\log_a(b)}$

where $a$ is any positive base. For example, $\log_3(9) = \frac{\ln(9)}{\ln(3)} = \frac{2.197}{1.099} = 2$, which checks out since $3^2 = 9$. This formula is especially useful when the base isn't 10 or $e$.

Properties and applications of logarithms

These properties turn multiplication, division, and exponentiation inside a log into addition, subtraction, and scalar multiplication outside. They're the main tools you'll use to expand, condense, and solve logarithmic expressions.

• Product rule: $\log_b(MN) = \log_b(M) + \log_b(N)$
  • $\log_2(8 \cdot 4) = \log_2(8) + \log_2(4) = 3 + 2 = 5$
• Quotient rule: $\log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N)$
  • $\log_3\left(\frac{81}{9}\right) = \log_3(81) - \log_3(9) = 4 - 2 = 2$
• Power rule: $\log_b(M^n) = n \log_b(M)$
  • $\log_2(16^3) = 3 \log_2(16) = 3 \cdot 4 = 12$
• Zero exponent rule: $\log_b(1) = 0$ for any valid base, because $b^0 = 1$
• Identity rule: $\log_b(b) = 1$ because $b^1 = b$

Solving logarithmic equations follows a consistent process:

1. Isolate the logarithm on one side of the equation.
2. Convert to exponential form (or apply the exponential function with the same base to both sides).
3. Solve the resulting equation.
4. Check your answer. The argument of a log must be positive, so plug your solution back in and reject any value that makes the argument zero or negative.

Graphing and analysis of logarithms

The parent function $f(x) = \log_b(x)$ (where $b > 0$ and $b \neq 1$) has these key features:

• Domain: $(0, \infty)$ — you can only take the log of positive numbers
• Range: $(-\infty, \infty)$
• Vertical asymptote: $x = 0$ (the graph approaches the y-axis but never touches it)
• x-intercept: always at $(1, 0)$ because $\log_b(1) = 0$ for any base
• Behavior: increasing if $b > 1$, decreasing if $0 < b < 1$

Since $f(x) = \log_b(x)$ is the inverse of $g(x) = b^x$, their graphs are reflections of each other across the line $y = x$.

Transformations follow the same rules as other function families:

• Vertical shift: $f(x) = \log_b(x) + k$ shifts the graph up by $k$ units
• Horizontal shift: $f(x) = \log_b(x - h)$ shifts the graph right by $h$ units (and moves the vertical asymptote to $x = h$)
• Vertical stretch/compression: $f(x) = a\log_b(x)$
  • Stretches vertically by a factor of $|a|$ if $|a| > 1$
  • Compresses vertically if $0 < |a| < 1$
• Reflection: $f(x) = -\log_b(x)$ reflects the graph over the x-axis

Watch the horizontal shift carefully: it changes the domain and the asymptote location. For $f(x) = \log_2(x - 3)$, the domain becomes $(3, \infty)$ and the asymptote moves to $x = 3$.

Real-world logarithmic problem solving

Logarithmic scales compress huge ranges of values into manageable numbers. That's why they appear in so many measurement systems.

Richter scale (earthquake magnitude): $M = \log\left(\frac{I}{I_0}\right)$ where $M$ is magnitude, $I$ is earthquake intensity, and $I_0$ is a reference intensity. An earthquake with intensity 1000 times the reference has magnitude $\log(1000) = 3$. Each whole number increase in magnitude represents a 10-fold increase in intensity.

Decibel scale (sound intensity): $\beta = 10\log\left(\frac{I}{I_0}\right)$ where $\beta$ is the sound level in decibels, $I$ is sound intensity, and $I_0$ is the reference intensity (threshold of hearing). A sound 100 times the reference intensity has a level of $10\log(100) = 20$ dB. Because of the factor of 10 out front, every 10 dB increase represents a 10-fold increase in intensity.

pH scale (acidity): $pH = -\log[H^+]$ where $[H^+]$ is the hydrogen ion concentration in moles per liter. The negative sign means higher concentrations give lower pH values (more acidic).

Exponential growth and decay: $A(t) = A_0 e^{kt}$ where $A(t)$ is the amount at time $t$, $A_0$ is the initial amount, and $k$ is the growth ($k > 0$) or decay ($k < 0$) constant. Logarithms let you solve for $t$ or $k$ by isolating the exponent.

• Half-life formula: $t_{1/2} = \frac{\ln(0.5)}{k}$
• Example: A radioactive substance with a half-life of 10 days has decay constant $k = \frac{\ln(0.5)}{10} \approx -0.0693$ per day. To find how long until only 25% remains, you'd solve $0.25 = e^{-0.0693t}$, giving $t = \frac{\ln(0.25)}{-0.0693} = 20$ days (exactly two half-lives, which makes sense).