Logarithmic Functions

📈 Logarithmic Functions

Part 1 of 7 — Logs as the Inverse of Exponentials

Topics in This Part

🔑 Key Concept: A logarithm answers one question — "what exponent turns the base into this number?" In AP Precalculus, the logarithmic function $y=\log_b x$ is built as the inverse of the exponential function $y=b^x$, and almost everything else follows from that single idea.

Logs Undo Exponentials

An exponential function takes an exponent and returns a value. A logarithm reverses that — it takes a value and returns the exponent.

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

We require the base $b>0$ and $b\ne 1$, and the input of a log must be positive.

The Two Forms Side by Side

🔑 Key Idea: Because $\log_b x$ and $b^x$ are inverses, their graphs are reflections across the line $y=x$, and they swap domain and range: $b^x$ has range $(0,\infty)$, so $\log_b x$ has domain $(0,\infty)$.

Concept Check 🎯

Evaluate by Hand 🧮

Ask yourself "what exponent turns the base into the input?"

1) $\log_2 16 = \,?$ 2) $\log_5 125 = \,?$ 3) $\log_3 \dfrac{1}{9} = \,?$

Two Special Bases

Two bases appear so often they get their own shorthand:

• Common log — base $10$, written $\log x$ (no base shown means base $10$).
• Natural log — base $e \approx 2.718$, written $\ln x$.

$\log x = \log_{10} x \qquad\qquad \ln x = \log_e x$

Quick values to know

💡 For any base, $\log_b 1 = 0$ (since $b^0=1$) and $\log_b b = 1$ (since $b^1 = b$). These two facts save time on every problem.

Match the Notation 🔽

Identify each value. Use the inverse relationship and the special-base facts.

📈 Logarithmic Functions

Part 2 of 7 — The Graph, Domain & Asymptote

🔑 The Idea: Reflecting $y=b^x$ across $y=x$ produces $y=\log_b x$. That reflection turns the exponential's horizontal asymptote into the log's vertical asymptote, and tells you the domain at a glance.

The Shape of $y=\log_b x$ (for $b>1$)

Because $y=\log_b x$ is the reflection of $y=b^x$ across $y=x$:

For $b>1$ the log is increasing and concave down: it climbs forever but more and more slowly.

⚠️ Domain trap: $\log_b x$ is only defined for $x>0$. You can never take the log of $0$ or a negative number — the graph hugs the line $x=0$ but never touches or crosses it.

End Behavior & Rate

As $x \to 0^+$, the curve plunges: $\log_b x \to -\infty$ (that's the vertical asymptote). As $x \to \infty$, it rises without bound but ever more slowly: $\log_b x \to \infty$.

Why "slowly"?

To increase $\log_2 x$ by just $1$, you must double $x$:

$\log_2 8 = 3,\quad \log_2 16 = 4,\quad \log_2 32 = 5$

Each step up costs twice as much input. This slow, decelerating growth is exactly why logs are used to compress huge ranges (decibels, pH, Richter scale).

💡 A logarithmic function grows slower than every positive power of $x$ — slower than $\sqrt{x}$, slower than $x^{0.01}$ — yet it still heads to $+\infty$.

Concept Check 🎯

Find the Domain Boundary 🧮

The argument of a log must be positive. Find the value of $x$ where each function's vertical asymptote sits (solve argument $=0$).

1) $f(x)=\log_3(x-4)$. Asymptote at $x = \,?$ 2) $g(x)=\ln(x+2)$. Asymptote at $x = \,?$ 3) $h(x)=\log(2x-6)$. Asymptote at $x = \,?$

Read the Transformation 🔽

Compare each function to the parent $y=\log_2 x$ and identify how the graph shifts.

📈 Logarithmic Functions

Part 3 of 7 — The Three Log Properties

🔑 Why they exist: Logs turn exponents into multipliers, so the exponent rules ($b^m b^n=b^{m+n}$, etc.) become log rules. These three properties let you expand, condense, and ultimately solve.

The Product, Quotient & Power Rules

For $b>0,\,b\ne 1$ and positive $M,N$:

$\textbf{Product: }\;\log_b(MN) = \log_b M + \log_b N$ $\textbf{Quotient: }\;\log_b\!\left(\frac{M}{N}\right) = \log_b M - \log_b N$ $\textbf{Power: }\;\log_b\!\left(M^p\right) = p\,\log_b M$

Where they come from

Multiplying numbers adds their exponents, so the log of a product adds the logs. Raising to a power multiplies the exponent, so the log of a power multiplies by that power.

⚠️ Top traps: $\log_b(M+N) \ne \log_b M + \log_b N$, and $\dfrac{\log_b M}{\log_b N} \ne \log_b\!\left(\dfrac{M}{N}\right)$. The rules apply to products, quotients, and powers — never to sums or differences inside one log.

Worked Example — Expanding

Write $\log_b \dfrac{x^3 y}{z}$ as a sum/difference of simpler logs.

$\log_b \frac{x^3 y}{z} = \log_b(x^3 y) - \log_b z \quad\text{(quotient)}$ $= \log_b x^3 + \log_b y - \log_b z \quad\text{(product)}$ $= 3\log_b x + \log_b y - \log_b z \quad\text{(power)}$

Worked Example — Condensing

Write $2\log x + \log 3 - \log y$ as a single logarithm.

$2\log x + \log 3 - \log y = \log x^2 + \log 3 - \log y \quad\text{(power)}$ $= \log(3x^2) - \log y \quad\text{(product)}$ $= \log\!\left(\frac{3x^2}{y}\right) \quad\text{(quotient)}$

✅ Check: Coefficients become exponents; pluses become multiplication; minuses become division.

Concept Check 🎯

Apply One Rule 🧮

Use the log properties. Enter the missing number.

1) $\log_2(8\cdot 4) = \log_2 8 + \log_2 4 = \,?$ 2) $\log_5 \dfrac{125}{25} = \log_5 125 - \log_5 25 = \,?$ 3) $\log_3 9^4 = 4\log_3 9 = \,?$

Expand or Condense 🔽

📈 Logarithmic Functions

Part 4 of 7 — Change of Base

🔑 The Problem: Your calculator only has $\log$ (base $10$) and $\ln$ (base $e$). The change-of-base formula lets you compute a log of any base using those two buttons.

The Change-of-Base Formula

For any valid bases $a,b$ and positive $x$:

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

In practice you pick $a=10$ or $a=e$:

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

Worked Example: $\log_2 50$

$\log_2 50 = \frac{\ln 50}{\ln 2} = \frac{3.912}{0.693} \approx 5.644$

Sanity check: $2^5 = 32$ and $2^6 = 64$, and $50$ is between them, so the answer should be between $5$ and $6$. ✓

💡 Memory aid: "new over new" — put the log of the argument on top and the log of the old base on the bottom, both in your new (calculator) base.

Change of Base for Exact Values

Change of base also gives exact answers when the numbers cooperate.

Example: $\log_8 32$

Convert to base $2$ (since $8=2^3$ and $32=2^5$):

$\log_8 32 = \frac{\log_2 32}{\log_2 8} = \frac{5}{3}$

Check: $8^{5/3} = \left(8^{1/3}\right)^5 = 2^5 = 32$ ✓

Example: $\log_9 27$

$\log_9 27 = \frac{\log_3 27}{\log_3 9} = \frac{3}{2}$

✅ Choosing a base where both numbers are clean powers turns a messy log into a simple fraction.

Concept Check 🎯

Exact Change of Base 🧮

Rewrite each in a common base, then enter the exact value (a whole number or a fraction like $3/2$).

1) $\log_4 16 = \,?$ 2) $\log_8 2 = \,?$ (fraction ok) 3) $\log_{27} 9 = \,?$ (fraction ok)

Set Up the Quotient 🔽

Choose the correct change-of-base setup (use base $e$).

📈 Logarithmic Functions

Part 5 of 7 — Solving Exponential & Logarithmic Equations

🔑 The Strategy: Logs and exponentials are inverses, so they undo each other. To free a variable stuck in an exponent, take a log; to free a variable stuck in a log, exponentiate.

Solving Exponential Equations

When the variable is in the exponent, take a log of both sides and use the power rule to bring it down.

Example: $3^x = 20$

$\ln 3^x = \ln 20 \;\Rightarrow\; x\ln 3 = \ln 20 \;\Rightarrow\; x = \frac{\ln 20}{\ln 3} \approx 2.727$

Example: $5\cdot 2^x = 80$

First isolate the exponential: $2^x = 16$. Then $x = \log_2 16 = 4$.

⚠️ Isolate first. Get the exponential term alone before taking a log. You can't bring the exponent down while a coefficient is still attached by multiplication.

Solving Logarithmic Equations

When the variable is inside a log, rewrite in exponential form (or condense to a single log first).

Example: $\log_2(x-1) = 5$

Rewrite as exponential: $x-1 = 2^5 = 32 \Rightarrow x = 33$.

Example: $\log x + \log(x-3) = 1$

Condense: $\log\big(x(x-3)\big) = 1 \Rightarrow x(x-3) = 10^1 = 10$. $x^2 - 3x - 10 = 0 \Rightarrow (x-5)(x+2) = 0 \Rightarrow x = 5 \text{ or } x=-2$

⚠️ Check for extraneous solutions! A log's argument must be positive. $x=-2$ makes $\log(-2)$ undefined, so we reject it. Only $x=5$ is valid.

Concept Check 🎯

Solve It 🧮

Solve each equation. Enter the (valid) value of $x$.

1) $2^x = 64 \;\Rightarrow\; x = \,?$ 2) $\log_5(x+1) = 3 \;\Rightarrow\; x = \,?$ 3) $\ln x = 0 \;\Rightarrow\; x = \,?$

Choose the First Move 🔽

For each equation, pick the correct first step.

📈 Logarithmic Functions

Part 6 of 7 — Logarithmic Models & Semi-Log Plots

🔑 The AP Connection: Logarithmic functions model the inverse of exponential growth — data that rises fast then levels off. And plotting exponential data on a semi-log scale turns its curve into a straight line you can analyze.

Where Logarithmic Models Fit

A logarithmic model $y = a + b\ln x$ (or with $\log$) is appropriate when the output grows quickly at first and then flattens — rapid early change, diminishing returns later.

Real-world log scales

💡 Order-of-magnitude rule: because these are base-$10$ logs, each unit increase means $\times 10$. A pH of $4$ is $10$ times more acidic than pH $5$; a magnitude-$6$ quake has $10$ times the amplitude of a magnitude-$5$ quake.

Semi-Log Plots Linearize Exponentials

Take an exponential model $y = a\cdot b^x$ and apply $\log$ to both sides:

$\log y = \log(a\cdot b^x) = \log a + x\log b$

Let $Y = \log y$. Then

$Y = (\log b)\,x + \log a$

That's a straight line in $x$ and $Y$! So if you plot $\log y$ (vertical) against $x$ (horizontal) — a semi-log plot — exponential data becomes linear.

🔑 Key takeaway: Exponential data is linear under a logarithmic transformation of the output. A straight semi-log plot is the fingerprint of exponential growth.

Concept Check 🎯

Read a Log Scale 🧮

Use the order-of-magnitude rule (each base-$10$ unit $= \times 10$).

1) Solution A has pH $3$, solution B has pH $6$. A is how many times more acidic than B? Enter the number. 2) A semi-log fit gives $\log y = 0.5x + 2$. What is $a = 10^{\text{intercept}}$? 3) Same fit: what is the base $b = 10^{\text{slope}}$? Use $10^{0.5}\approx 3.16$ and round to two decimals.

Interpret the Semi-Log Line 🔽

A data set is plotted as $\log y$ versus $x$ and fits the line $\log y = 0.3x + 1$.

📈 Logarithmic Functions

Part 7 of 7 — Mixed Mastery & Exit Quiz

You can now (1) read logs as inverses of exponentials, (2) graph them and state domain/asymptote, (3) apply the three properties, (4) change base, (5) solve exp/log equations, and (6) interpret log models and semi-log plots. Let's pull it together.

Quick Reference

⚠️ Three things never to forget: a log's argument is always positive, $\log_b(M+N)\ne\log_b M+\log_b N$, and check for extraneous solutions whenever you condense logs.

Mixed Practice 🎯

Mixed Drill 🧮

1) Evaluate $\log_{16} 4$. (fraction ok) 2) Solve $10^{x} = 1000 \Rightarrow x = \,?$ 3) Vertical asymptote of $f(x)=\ln(x-7)$ is at $x = \,?$

Exit Quiz ✅

Answer all three to finish the lesson.