Logarithmic Functions

Unlock the inverse of exponential functions

If you have ever tried to solve an equation like $2^x = 10$ and felt stuck, you are not alone. With what you know so far, you can express 10 as a power of 2 only through trial and error. Is the answer 3? Well, $2^3 = 8$, which is too small. What about 4? That gives $2^4 = 16$, which is too big. The answer is somewhere between 3 and 4, but how do you find it exactly?

This is where logarithms come to the rescue. Logarithms are specifically designed to answer the question: “What exponent do I need?” They undo exponential functions just like subtraction undoes addition or division undoes multiplication. Once you understand this simple idea, logarithms become much less mysterious.

In this chapter, we are going to explore logarithmic functions, learn their properties, and discover how they help us solve problems that would otherwise be nearly impossible. You will see that logarithms are not some arcane mathematical concept reserved for specialists; they show up in earthquake measurements, sound levels, pH scales, and even how we perceive brightness and loudness.

Core Concepts

What Is a Logarithm?

A logarithm answers the question: “What power must I raise the base to in order to get this number?”

The expression $\log_b(x)$ asks: “What exponent do I need to put on $b$ to get $x$?”

Here is the fundamental relationship:

$$y = \log_b(x) \quad \text{means} \quad b^y = x$$

Let us read this carefully: “$y$ equals log base $b$ of $x$” means “$b$ raised to the power $y$ equals $x$.”

For example:

  • $\log_2(8) = 3$ because $2^3 = 8$
  • $\log_3(81) = 4$ because $3^4 = 81$
  • $\log_{10}(1000) = 3$ because $10^3 = 1000$

Every logarithm is simply asking an exponent question. When you see $\log_2(8)$, just ask yourself: “2 to what power gives me 8?” The answer is 3.

Logarithms as Inverses of Exponentials

If you studied exponential functions, you learned that they are one-to-one: each input produces a unique output, and each output comes from a unique input. This means exponential functions have inverses, and those inverses are logarithmic functions.

The exponential function $f(x) = b^x$ and the logarithmic function $g(x) = \log_b(x)$ are inverses of each other. They “undo” each other:

$$\log_b(b^x) = x \quad \text{for all real numbers } x$$

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

Think of it this way: if the exponential function takes you “up” from an exponent to a result, the logarithm takes you “back down” from the result to the exponent.

Common Logarithm and Natural Logarithm

While you can have a logarithm with any positive base (other than 1), two bases are used so frequently that they have special notations:

The Common Logarithm (base 10): $$\log(x) = \log_{10}(x)$$

When you see “log” without a base written, it almost always means base 10. This is the logarithm built into most basic calculators. Base 10 is natural for us because we use a decimal number system.

The Natural Logarithm (base $e$): $$\ln(x) = \log_e(x)$$

The natural logarithm uses the special number $e \approx 2.71828$ as its base. Just as $e$ appears naturally in continuous growth and calculus, the natural logarithm appears naturally whenever we work with exponential growth or decay. The notation “ln” comes from the Latin logarithmus naturalis.

Both of these are available on scientific calculators, usually labeled “log” and “ln.”

Domain Restrictions

Here is something crucial to understand: you can only take the logarithm of a positive number.

Why? Think about what $\log_2(x)$ is asking: “2 to what power gives $x$?” If $x = 0$, there is no power you can raise 2 to that gives 0 (powers of positive numbers are always positive). If $x$ is negative, say $x = -8$, there is no real number power that makes $2^? = -8$.

Therefore, for any logarithmic function $f(x) = \log_b(x)$:

  • Domain: $x > 0$ (all positive real numbers)
  • Range: All real numbers

This is the opposite of exponential functions, which have domain all real numbers and range all positive numbers. As inverses, they swap domains and ranges.

Notation and Terminology

Term Symbol Meaning
Logarithm $\log_b(x)$ The exponent you need on base $b$ to get $x$
Base $b$ The number being raised to a power (must be positive, not 1)
Argument $x$ The number inside the logarithm (must be positive)
Common logarithm $\log(x)$ Logarithm base 10; asks “10 to what power gives $x$?”
Natural logarithm $\ln(x)$ Logarithm base $e$; asks “$e$ to what power gives $x$?”
Exponential form $b^y = x$ The equivalent exponential statement
Logarithmic form $y = \log_b(x)$ The equivalent logarithmic statement

Converting Between Exponential and Logarithmic Form

One of the most important skills with logarithms is converting between the two forms. This is not a new calculation; it is just two ways of writing the same relationship.

$$\text{Logarithmic form: } y = \log_b(x) \quad \Longleftrightarrow \quad \text{Exponential form: } b^y = x$$

The base stays the base. The logarithm equals the exponent. The argument is the result of the exponential.

Example 1: Converting to Exponential Form

Convert each logarithmic equation to exponential form:

(a) $\log_5(125) = 3$

(b) $\log_2(1/8) = -3$

(c) $\ln(e^4) = 4$

Solutions:

(a) The base is 5, the logarithm (exponent) is 3, and the argument (result) is 125. $$5^3 = 125$$

(b) The base is 2, the exponent is $-3$, and the result is $1/8$. $$2^{-3} = \frac{1}{8}$$

(c) The base is $e$, the exponent is 4, and the result is $e^4$. $$e^4 = e^4$$

This last one might seem trivial, but it illustrates that $\ln(e^x) = x$ for any $x$.

Example 2: Converting to Logarithmic Form

Convert each exponential equation to logarithmic form:

(a) $4^3 = 64$

(b) $10^{-2} = 0.01$

(c) $e^0 = 1$

Solutions:

(a) The base is 4, the exponent is 3, and the result is 64. $$\log_4(64) = 3$$

(b) The base is 10, the exponent is $-2$, and the result is 0.01. $$\log_{10}(0.01) = -2 \quad \text{or equivalently} \quad \log(0.01) = -2$$

(c) The base is $e$, the exponent is 0, and the result is 1. $$\ln(1) = 0$$

This confirms that the natural log of 1 is 0, which makes sense: any number to the power 0 equals 1.

Evaluating Logarithms

To evaluate a logarithm without a calculator, convert it to the question it is really asking, then find the answer.

Example 3: Evaluating Logarithms by Inspection

Evaluate each logarithm:

(a) $\log_3(27)$

(b) $\log_4(1/16)$

(c) $\log_{100}(10)$

Solutions:

(a) Ask: “3 to what power equals 27?” Since $3^1 = 3$, $3^2 = 9$, $3^3 = 27$, the answer is 3. $$\log_3(27) = 3$$

(b) Ask: “4 to what power equals $1/16$?” We know $4^2 = 16$, so $4^{-2} = 1/16$. $$\log_4(1/16) = -2$$

(c) Ask: “100 to what power equals 10?” Since $100 = 10^2$, we need $(10^2)^? = 10^1$. Using the power rule: $10^{2 \cdot ?} = 10^1$, so $2 \cdot ? = 1$, meaning $? = 1/2$. $$\log_{100}(10) = \frac{1}{2}$$

You can verify: $100^{1/2} = \sqrt{100} = 10$. Correct!

Example 4: Evaluating Common and Natural Logarithms

Evaluate each logarithm:

(a) $\log(10000)$

(b) $\log(0.001)$

(c) $\ln(e^5)$

(d) $\ln(\sqrt{e})$

Solutions:

(a) Ask: “10 to what power equals 10,000?” $10^4 = 10,000$, so $\log(10000) = 4$.

(b) Ask: “10 to what power equals 0.001?” $0.001 = \frac{1}{1000} = 10^{-3}$, so $\log(0.001) = -3$.

(c) By the inverse property, $\ln(e^5) = 5$. The natural log “undoes” the exponential with base $e$.

(d) $\sqrt{e} = e^{1/2}$, so $\ln(\sqrt{e}) = \ln(e^{1/2}) = \frac{1}{2}$.

Key Properties and Rules

Just as exponents have properties that let us simplify expressions, logarithms have corresponding properties. These properties follow directly from the exponent rules because logarithms are exponents.

The Product Rule

$$\log_b(MN) = \log_b(M) + \log_b(N)$$

The logarithm of a product is the sum of the logarithms.

Why does this work? If $\log_b(M) = p$ and $\log_b(N) = q$, then $b^p = M$ and $b^q = N$. So $MN = b^p \cdot b^q = b^{p+q}$, which means $\log_b(MN) = p + q = \log_b(M) + \log_b(N)$.

The Quotient Rule

$$\log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N)$$

The logarithm of a quotient is the difference of the logarithms.

Why? Using similar reasoning: $\frac{M}{N} = \frac{b^p}{b^q} = b^{p-q}$.

The Power Rule

$$\log_b(M^n) = n \cdot \log_b(M)$$

The logarithm of a power brings the exponent down as a multiplier.

Why? $M^n = (b^p)^n = b^{pn}$, so $\log_b(M^n) = pn = n \cdot \log_b(M)$.

Special Values

$$\log_b(1) = 0 \quad \text{because } b^0 = 1$$

$$\log_b(b) = 1 \quad \text{because } b^1 = b$$

Summary of Properties

Property Rule Example
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 $\log_b(M/N) = \log_b(M) - \log_b(N)$ $\log_3(81/9) = \log_3(81) - \log_3(9) = 4 - 2 = 2$
Power Rule $\log_b(M^n) = n \cdot \log_b(M)$ $\log_5(25^3) = 3 \cdot \log_5(25) = 3 \cdot 2 = 6$
Log of 1 $\log_b(1) = 0$ $\log_7(1) = 0$
Log of base $\log_b(b) = 1$ $\log_{10}(10) = 1$
Inverse properties $\log_b(b^x) = x$ and $b^{\log_b(x)} = x$ $\log_2(2^7) = 7$ and $5^{\log_5(12)} = 12$
Example 5: Expanding Logarithmic Expressions

Use logarithm properties to expand each expression:

(a) $\log_3(9x^2)$

(b) $\ln\left(\frac{x^3}{e^2}\right)$

(c) $\log\left(\frac{100\sqrt{y}}{z^4}\right)$

Solutions:

(a) Using the product rule, then the power rule: $$\log_3(9x^2) = \log_3(9) + \log_3(x^2) = 2 + 2\log_3(x)$$ (Since $\log_3(9) = 2$ because $3^2 = 9$.)

(b) Using the quotient rule, then the power rule: $$\ln\left(\frac{x^3}{e^2}\right) = \ln(x^3) - \ln(e^2) = 3\ln(x) - 2$$ (Since $\ln(e^2) = 2$.)

(c) First rewrite $\sqrt{y} = y^{1/2}$, then apply the rules: $$\log\left(\frac{100\sqrt{y}}{z^4}\right) = \log(100) + \log(y^{1/2}) - \log(z^4)$$ $$= 2 + \frac{1}{2}\log(y) - 4\log(z)$$

Example 6: Condensing Logarithmic Expressions

Write each expression as a single logarithm:

(a) $\log_2(x) + \log_2(x-1)$

(b) $3\ln(x) - \ln(y)$

(c) $2\log(5) + \log(4)$

Solutions:

(a) Using the product rule in reverse: $$\log_2(x) + \log_2(x-1) = \log_2[x(x-1)] = \log_2(x^2 - x)$$

(b) First apply the power rule, then the quotient rule: $$3\ln(x) - \ln(y) = \ln(x^3) - \ln(y) = \ln\left(\frac{x^3}{y}\right)$$

(c) Apply the power rule, then the product rule: $$2\log(5) + \log(4) = \log(5^2) + \log(4) = \log(25) + \log(4) = \log(25 \cdot 4) = \log(100) = 2$$

The Change of Base Formula

What if you need to calculate $\log_3(17)$ but your calculator only has buttons for $\log$ (base 10) and $\ln$ (base $e$)? The change of base formula lets you convert any logarithm to a different base:

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

This works for any base $c$, but the two most useful versions are:

$$\log_b(x) = \frac{\log(x)}{\log(b)} = \frac{\ln(x)}{\ln(b)}$$

Why does this work? Let $y = \log_b(x)$, so $b^y = x$. Take the logarithm (any base $c$) of both sides: $$\log_c(b^y) = \log_c(x)$$ $$y \cdot \log_c(b) = \log_c(x)$$ $$y = \frac{\log_c(x)}{\log_c(b)}$$

Example 7: Using the Change of Base Formula

Calculate $\log_3(17)$ to three decimal places.

Using common logarithms: $$\log_3(17) = \frac{\log(17)}{\log(3)} = \frac{1.2304…}{0.4771…} \approx 2.579$$

Using natural logarithms: $$\log_3(17) = \frac{\ln(17)}{\ln(3)} = \frac{2.8332…}{1.0986…} \approx 2.579$$

Check: $3^{2.579} \approx 17.0$. It works!

Both methods give the same answer, so use whichever is more convenient.

Graphing Logarithmic Functions

Since logarithmic functions are inverses of exponential functions, their graphs are reflections of each other across the line $y = x$.

The Basic Shape of $f(x) = \log_b(x)$

For any base $b > 1$:

  1. Domain: $x > 0$ (only positive inputs allowed)
  2. Range: All real numbers (the output can be anything)
  3. $x$-intercept: $(1, 0)$ because $\log_b(1) = 0$
  4. No $y$-intercept: The function is not defined at $x = 0$
  5. Vertical asymptote: The $y$-axis ($x = 0$). As $x$ approaches 0 from the right, the function decreases without bound.
  6. Key point: $(b, 1)$ because $\log_b(b) = 1$
  7. Behavior: Increases slowly as $x$ increases (concave down)

For $0 < b < 1$, the graph is reflected across the $x$-axis: it decreases as $x$ increases and approaches positive infinity as $x$ approaches 0.

Transformations

Just like other functions, logarithmic functions can be transformed:

Vertical shifts: $f(x) = \log_b(x) + k$

  • Shifts the graph up by $k$ units (if $k > 0$) or down (if $k < 0$)
  • The vertical asymptote stays at $x = 0$

Horizontal shifts: $f(x) = \log_b(x - h)$

  • Shifts the graph right by $h$ units (if $h > 0$) or left (if $h < 0$)
  • The vertical asymptote moves to $x = h$
  • The domain becomes $x > h$

Vertical stretches and reflections: $f(x) = a \cdot \log_b(x)$

  • If $|a| > 1$, the graph is stretched vertically
  • If $0 < |a| < 1$, the graph is compressed vertically
  • If $a < 0$, the graph is reflected across the $x$-axis

Horizontal reflections: $f(x) = \log_b(-x)$

  • Reflects across the $y$-axis
  • The domain becomes $x < 0$
Example 8: Graphing Transformed Logarithmic Functions

Describe the transformations applied to $f(x) = \ln(x)$ to obtain $g(x) = -2\ln(x - 3) + 1$, and identify the domain, vertical asymptote, and any intercepts.

Transformations:

  1. Horizontal shift right 3 units (from $x - 3$)
  2. Vertical stretch by factor of 2 (from the coefficient 2)
  3. Reflection across the $x$-axis (from the negative sign)
  4. Vertical shift up 1 unit (from the $+1$)

Domain: The argument must be positive: $x - 3 > 0$, so $x > 3$. Domain: $(3, \infty)$

Vertical asymptote: The original asymptote at $x = 0$ shifts right by 3. Vertical asymptote: $x = 3$

$x$-intercept: Set $g(x) = 0$: $$0 = -2\ln(x - 3) + 1$$ $$2\ln(x - 3) = 1$$ $$\ln(x - 3) = \frac{1}{2}$$ $$x - 3 = e^{1/2} = \sqrt{e}$$ $$x = 3 + \sqrt{e} \approx 4.649$$

$x$-intercept: $(3 + \sqrt{e}, 0) \approx (4.649, 0)$

$y$-intercept: None, since $x = 0$ is not in the domain.

Solving Logarithmic Equations

Logarithmic equations can be solved using the definition of logarithms and their properties.

Strategy 1: Convert to Exponential Form

If the equation has a single logarithm, isolate it and convert to exponential form.

Example 9: Solving by Converting to Exponential Form

Solve: $\log_2(x - 5) = 4$

Step 1: Convert to exponential form: $$2^4 = x - 5$$ $$16 = x - 5$$

Step 2: Solve for $x$: $$x = 21$$

Step 3: Check that the answer is in the domain. The argument was $x - 5$, which must be positive: $21 - 5 = 16 > 0$. Valid!

Answer: $x = 21$

Strategy 2: Use the One-to-One Property

If $\log_b(M) = \log_b(N)$, then $M = N$ (as long as both $M$ and $N$ are positive).

Example 10: Using Logarithm Properties to Solve Equations

Solve: $\log_3(x) + \log_3(x - 2) = 1$

Step 1: Use the product rule to combine: $$\log_3[x(x - 2)] = 1$$ $$\log_3(x^2 - 2x) = 1$$

Step 2: Convert to exponential form: $$3^1 = x^2 - 2x$$ $$3 = x^2 - 2x$$

Step 3: Solve the quadratic: $$x^2 - 2x - 3 = 0$$ $$(x - 3)(x + 1) = 0$$ $$x = 3 \quad \text{or} \quad x = -1$$

Step 4: Check both solutions in the original equation.

For $x = 3$: $\log_3(3) + \log_3(3 - 2) = \log_3(3) + \log_3(1) = 1 + 0 = 1$. Valid!

For $x = -1$: $\log_3(-1)$ is undefined (negative argument). Reject this solution.

Answer: $x = 3$

Example 11: Solving Exponential Equations Using Logarithms

Solve: $5^{2x-1} = 17$

This equation cannot be solved by the same-base method because 17 is not a nice power of 5. We use logarithms.

Step 1: Take the natural logarithm of both sides: $$\ln(5^{2x-1}) = \ln(17)$$

Step 2: Apply the power rule: $$(2x - 1) \cdot \ln(5) = \ln(17)$$

Step 3: Solve for $x$: $$2x - 1 = \frac{\ln(17)}{\ln(5)}$$ $$2x = \frac{\ln(17)}{\ln(5)} + 1$$ $$x = \frac{1}{2}\left(\frac{\ln(17)}{\ln(5)} + 1\right)$$

Step 4: Calculate numerically: $$x = \frac{1}{2}\left(\frac{2.833…}{1.609…} + 1\right) = \frac{1}{2}(1.760… + 1) = \frac{1}{2}(2.760…) \approx 1.380$$

Answer: $x \approx 1.380$

Check: $5^{2(1.380) - 1} = 5^{1.760} \approx 17$. Correct!

Real-World Applications

Logarithms appear in countless real-world contexts, often because they help us measure quantities that span huge ranges.

The Decibel Scale (Sound Intensity)

Sound intensity varies enormously, from a whisper to a jet engine. The decibel scale uses logarithms to compress this range:

$$\text{Loudness (dB)} = 10 \cdot \log\left(\frac{I}{I_0}\right)$$

Where $I$ is the sound intensity and $I_0$ is the threshold of human hearing.

  • A whisper: about 30 dB
  • Normal conversation: about 60 dB
  • Rock concert: about 110 dB
  • Pain threshold: about 130 dB

Each increase of 10 dB represents a tenfold increase in actual intensity, but our ears perceive it as roughly “twice as loud.”

The Richter Scale (Earthquakes)

Earthquake magnitudes are measured on a logarithmic scale:

$$M = \log\left(\frac{A}{A_0}\right)$$

Each whole number increase represents a tenfold increase in measured wave amplitude. In terms of energy released, each whole number increase represents about 31.6 times more energy.

So a magnitude 7 earthquake releases about 31.6 times more energy than a magnitude 6, and about 1000 times more energy than a magnitude 5.

The pH Scale (Acidity)

The pH of a solution measures its acidity:

$$\text{pH} = -\log[H^+]$$

Where $[H^+]$ is the hydrogen ion concentration in moles per liter.

  • pH 7 is neutral (pure water)
  • pH below 7 is acidic (lemon juice is about pH 2)
  • pH above 7 is basic (bleach is about pH 13)

Each unit change in pH represents a tenfold change in hydrogen ion concentration.

Radioactive Dating

When scientists determine the age of artifacts using carbon-14 dating, they use the logarithmic relationship:

$$t = \frac{\ln(N_0/N)}{k}$$

Where $N_0$ is the original amount of carbon-14, $N$ is the current amount, and $k$ is the decay constant.

Information Theory and Computing

The number of bits needed to represent $n$ different values is $\lceil\log_2(n)\rceil$ (the ceiling of log base 2). This is why computer scientists frequently use base-2 logarithms.

For example, to uniquely identify one item out of 1000 possibilities, you need $\log_2(1000) \approx 10$ bits.

Population and Investment Growth

From the compound interest formula, we can solve for time using logarithms:

$$t = \frac{\ln(A/P)}{r}$$

This tells us how long it takes for an investment to grow from $P$ to $A$ at continuous interest rate $r$.

Self-Test Problems

Problem 1: Convert $5^3 = 125$ to logarithmic form.

Show Answer

The base is 5, the exponent is 3, and the result is 125.

$$\log_5(125) = 3$$

Problem 2: Evaluate $\log_4(64)$ without a calculator.

Show Answer

Ask: “4 to what power equals 64?”

$4^1 = 4$, $4^2 = 16$, $4^3 = 64$.

$$\log_4(64) = 3$$

Problem 3: Use logarithm properties to expand $\log_2\left(\frac{x^3 y}{8}\right)$.

Show Answer

Apply the quotient rule, then the product rule, then the power rule:

$$\log_2\left(\frac{x^3 y}{8}\right) = \log_2(x^3 y) - \log_2(8)$$ $$= \log_2(x^3) + \log_2(y) - \log_2(8)$$ $$= 3\log_2(x) + \log_2(y) - 3$$

(Since $\log_2(8) = 3$ because $2^3 = 8$.)

Problem 4: Condense into a single logarithm: $\frac{1}{2}\ln(x) - 3\ln(y) + \ln(z)$

Show Answer

Apply the power rule first, then combine using product and quotient rules:

$$\frac{1}{2}\ln(x) - 3\ln(y) + \ln(z) = \ln(x^{1/2}) - \ln(y^3) + \ln(z)$$ $$= \ln(\sqrt{x}) + \ln(z) - \ln(y^3)$$ $$= \ln\left(\frac{z\sqrt{x}}{y^3}\right)$$

Problem 5: Solve: $\log_5(2x + 3) = 2$

Show Answer

Convert to exponential form: $$5^2 = 2x + 3$$ $$25 = 2x + 3$$ $$22 = 2x$$ $$x = 11$$

Check: $\log_5(2(11) + 3) = \log_5(25) = 2$. Correct!

Answer: $x = 11$

Problem 6: Solve: $3^{x+2} = 20$ (Give an exact answer and a decimal approximation.)

Show Answer

Take the natural logarithm of both sides: $$\ln(3^{x+2}) = \ln(20)$$ $$(x + 2)\ln(3) = \ln(20)$$ $$x + 2 = \frac{\ln(20)}{\ln(3)}$$ $$x = \frac{\ln(20)}{\ln(3)} - 2$$

Exact answer: $x = \frac{\ln(20)}{\ln(3)} - 2$

Decimal approximation: $$x = \frac{2.996…}{1.099…} - 2 \approx 2.727 - 2 = 0.727$$

Answer: $x = \frac{\ln(20)}{\ln(3)} - 2 \approx 0.727$

Problem 7: The pH of a solution is given by $\text{pH} = -\log[H^+]$. If a solution has a pH of 4.5, what is the hydrogen ion concentration $[H^+]$?

Show Answer

Start with the equation: $$4.5 = -\log[H^+]$$ $$-4.5 = \log[H^+]$$

Convert to exponential form: $$[H^+] = 10^{-4.5}$$

Calculate: $$[H^+] = 10^{-4.5} \approx 3.16 \times 10^{-5} \text{ mol/L}$$

Answer: $[H^+] \approx 3.16 \times 10^{-5}$ moles per liter

Summary

  • A logarithm answers the question “What exponent gives this result?” The equation $y = \log_b(x)$ means $b^y = x$.

  • Logarithms are inverses of exponential functions. They “undo” exponentiation: $\log_b(b^x) = x$ and $b^{\log_b(x)} = x$.

  • The common logarithm ($\log$) uses base 10. The natural logarithm ($\ln$) uses base $e \approx 2.718$.

  • Domain restriction: You can only take the logarithm of a positive number. The domain of $f(x) = \log_b(x)$ is $x > 0$.

  • Logarithm properties mirror exponent rules:

    • Product rule: $\log_b(MN) = \log_b(M) + \log_b(N)$
    • Quotient rule: $\log_b(M/N) = \log_b(M) - \log_b(N)$
    • Power rule: $\log_b(M^n) = n \cdot \log_b(M)$
  • The change of base formula lets you convert between bases: $\log_b(x) = \frac{\ln(x)}{\ln(b)} = \frac{\log(x)}{\log(b)}$.

  • Graphs of logarithmic functions have a vertical asymptote, pass through $(1, 0)$, and increase slowly. They are reflections of exponential graphs across the line $y = x$.

  • Transformations work similarly to other functions: $f(x) = a \cdot \log_b(x - h) + k$ shifts horizontally by $h$, vertically by $k$, and stretches or reflects based on $a$.

  • To solve logarithmic equations, either convert to exponential form or use the one-to-one property. Always check that solutions give positive arguments.

  • Real-world applications include the decibel scale (sound), Richter scale (earthquakes), pH scale (acidity), carbon dating, and calculating investment growth time.

  • Logarithms help us work with quantities that span enormous ranges by compressing them into manageable numbers. This is why they appear throughout science, engineering, and finance.