10.3 Evaluate and Graph Logarithmic Functions

Exponential and logarithmic expressions are powerful tools for modeling real-world phenomena. They describe relationships where quantities change at rates proportional to their current values, like compound interest or radioactive decay.

These functions are inverses of each other: exponential form expresses $y$ in terms of $x$, while logarithmic form expresses $x$ in terms of $y$. Understanding their properties and graphs is essential for solving problems in science, finance, and engineering.

Exponential and Logarithmic Expressions

Exponential and logarithmic conversions

A logarithm answers the question: "What exponent do I need?" The two forms below say the exact same thing, just written differently.

• Exponential form $b^x = y$ states that base $b$ raised to exponent $x$ gives result $y$. For example, $2^3 = 8$.
• Logarithmic form $\log_b(y) = x$ asks "what power of $b$ gives $y$?" For example, $\log_2(8) = 3$ because $2^3 = 8$.

To convert between the two forms, remember this pattern:

$b^x = y \quad \Longleftrightarrow \quad \log_b(y) = x$

They are inverse operations, just like multiplication and division undo each other.

Two special logarithms come up constantly:

• Common logarithm $\log(x)$ has base 10. When you see $\log$ with no base written, assume base 10. For example, $\log(100) = 2$ because $10^2 = 100$.
• Natural logarithm $\ln(x)$ has base $e \approx 2.71828$. For example, $\ln(e) = 1$ because $e^1 = e$.

Evaluation of logarithmic functions

These four properties let you break apart, combine, and simplify logarithmic expressions. They work for any valid base $b$.

• Product rule: $\log_b(M \cdot N) = \log_b(M) + \log_b(N)$ A log of a product becomes a sum of logs. For example, $\log_2(4 \cdot 8) = \log_2(4) + \log_2(8) = 2 + 3 = 5$.

• Quotient rule: $\log_b\!\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N)$ A log of a quotient becomes a difference of logs. For example, $\log_3\!\left(\frac{81}{9}\right) = \log_3(81) - \log_3(9) = 4 - 2 = 2$.

• Power rule: $\log_b(M^n) = n \cdot \log_b(M)$ An exponent inside the log can be pulled out front as a multiplier. For example, $\log_2(8^3) = 3 \cdot \log_2(8) = 3 \cdot 3 = 9$.

• Change of base formula: $\log_b(x) = \frac{\log_a(x)}{\log_a(b)}$ This lets you rewrite any logarithm in terms of a base your calculator can handle (base 10 or base $e$). For example, $\log_5(20) = \frac{\ln(20)}{\ln(5)} \approx \frac{3.0}{1.609} \approx 1.861$.

These properties are essential for simplifying expressions and solving logarithmic equations.

Graphing and Solving Logarithmic Functions

Graphs of logarithmic functions

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

• Domain: $(0, \infty)$. The input must be positive; $\log_b(x)$ is undefined for $x \leq 0$.
• Range: $(-\infty, \infty)$. The output can be any real number.
• Vertical asymptote: $x = 0$. The graph approaches the y-axis but never touches it.
• Key point: The graph always passes through $(1, 0)$ because $\log_b(1) = 0$ for every base.
• Direction: If $b > 1$, the function increases (e.g., $\log_2(x)$). If $0 < b < 1$, the function decreases (e.g., $\log_{1/2}(x)$).

Transformations shift, stretch, or flip the parent graph:

• Vertical shift: $f(x) = \log_b(x) + k$ moves the graph up $k$ units (down if $k < 0$). Example: $\log_2(x) + 1$ shifts up 1 unit.
• Horizontal shift: $f(x) = \log_b(x - h)$ moves the graph right $h$ units (left if $h < 0$). This also moves the vertical asymptote to $x = h$. Example: $\log_2(x - 3)$ shifts right 3 units, with asymptote at $x = 3$.
• Vertical stretch/compression: $f(x) = a \cdot \log_b(x)$ stretches vertically by factor $|a|$ when $|a| > 1$, compresses when $0 < |a| < 1$. Example: $2\log_2(x)$ stretches by a factor of 2.
• Reflection over the x-axis: $f(x) = -\log_b(x)$ flips the graph upside down. Example: $-\log_2(x)$.

Equations with logarithms

Here's a step-by-step approach for solving logarithmic equations:

1. Isolate the logarithm on one side of the equation. Example: $\log_3(x) + 2 = 5$ becomes $\log_3(x) = 3$.

2. Convert to exponential form to "undo" the logarithm. $\log_3(x) = 3$ becomes $3^3 = x$, so $x = 27$.

3. If two logs with the same base are equal, set their arguments equal. $\log_3(x) = \log_3(27)$ means $x = 27$.

4. If the bases differ, use the change of base formula to rewrite both sides with a common base, then solve.

5. Always check your answer. Plug it back in to make sure the argument of every logarithm is positive. Any solution that makes an argument zero or negative is extraneous and must be thrown out.

Real-world logarithmic applications

Logarithmic scales are used whenever a quantity spans a huge range of values. Each step on the scale represents a multiplication, not an addition.

• Richter scale: $M = \log_{10}\!\left(\frac{I}{I_0}\right)$ measures earthquake magnitude. An earthquake with intensity $I = 10{,}000 \cdot I_0$ has magnitude $M = \log_{10}(10{,}000) = 4$. Each increase of 1 on the scale means 10 times more intense.
• Decibel scale: $\beta = 10 \cdot \log_{10}\!\left(\frac{I}{I_0}\right)$ measures sound intensity in decibels. A sound with $I = 100 \cdot I_0$ has level $\beta = 10 \cdot \log_{10}(100) = 20$ dB.
• pH scale: $\text{pH} = -\log_{10}([H^+])$ measures how acidic or basic a solution is. A solution with $[H^+] = 10^{-7}$ mol/L has $\text{pH} = 7$ (neutral). Lower pH means more acidic; higher pH means more basic.

Exponential Growth and Decay

• Exponential growth is modeled by $f(t) = A \cdot e^{kt}$ where $A$ is the initial amount, $k > 0$ is the growth rate, and $t$ is time. The quantity increases faster and faster over time.
• Exponential decay is modeled by $f(t) = A \cdot e^{-kt}$ where $k > 0$. The quantity decreases, approaching zero but never reaching it.

Logarithms are the tool you use to solve for time in these models. For example, to find when a population doubles, you set $2A = A \cdot e^{kt}$, cancel $A$, and take the natural log of both sides:

$\ln(2) = kt \quad \Longrightarrow \quad t = \frac{\ln(2)}{k}$

This gives you the doubling time. The same approach works for half-life problems: set $\frac{1}{2}A = A \cdot e^{-kt}$ and solve to get $t = \frac{\ln(2)}{k}$.