Logarithmic Functions

🪵 Logarithmic Functions

Part 1 of 5 — What a Logarithm Really Is

Topics in This Part

🔑 Key Concept: A logarithm answers one question — "What exponent do I put on the base to get this number?" Everything in this lesson grows from that single idea.

Logarithms Undo Exponents

You already know how to read $2^3 = 8$ — "$2$ to the power $3$ equals $8$."

A logarithm flips that around. It starts with the answer ($8$) and asks for the exponent:

$\log_2 8 = 3 \quad\Longleftrightarrow\quad 2^3 = 8$

Read $\log_2 8 = 3$ as "the log, base $2$, of $8$ is $3$" — meaning $2$ raised to the $3$ gives $8$.

🔑 Key Idea: A log is an exponent. The expression $\log_b x$ literally equals the power you raise $b$ to in order to reach $x$.

The Definition

This is the most important line in the whole lesson — memorize it:

$\log_b x = y \quad\Longleftrightarrow\quad b^y = x \qquad (b > 0,\; b \ne 1,\; x > 0)$

• $b$ is the base (same base in both forms)
• $y$ is the exponent — and also the value of the log
• $x$ is the result (the number inside the log, called the argument)

Switching Forms

💡 Trick to remember: the base of the log becomes the base of the power, and the answer of the log becomes the exponent. The argument is left over on the other side.

Switch Between Forms 🔽

Match each logarithm to its equivalent exponential equation.

Evaluating a Logarithm

To find $\log_b x$ by hand, ask: "$b$ to what power gives $x$?"

Example: $\log_2 32$

$2$ to the power what equals $32$? Since $2^5 = 32$, the answer is $\boxed{5}$.

Example: $\log_3 \frac{1}{9}$

$3$ to the power what equals $\frac{1}{9}$? Since $3^{-2} = \frac{1}{9}$, the answer is $\boxed{-2}$.

Two Logs You Should Know Instantly

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

⚠️ Watch out: the argument of a log must be positive. There is no real value for $\log_2(-8)$ or $\log_2 0$ — no power of $2$ is ever negative or zero.

Evaluate Each Log 🧮

Ask "the base to what power gives the argument?" Enter the value.

1) $\log_2 64 = \,?$ 2) $\log_5 125 = \,?$ 3) $\log_4 \dfrac{1}{16} = \,?$

Common Logs and Natural Logs

Two bases are so common they get shorthand notation:

When you see $\log x$ with no base, assume base $10$. When you see $\ln x$, the base is the constant $e$.

$\log 100 = 2 \quad (10^2 = 100) \qquad\qquad \ln e = 1 \quad (e^1 = e)$

💡 Your calculator has dedicated $\boxed{\log}$ and $\boxed{\ln}$ keys — that's why these two bases are special. We'll use them in Part 4 to evaluate any base.

Concept Check 🎯

🪵 Logarithmic Functions

Part 2 of 5 — The Graph & Its Inverse

🔑 The Idea: The logarithm function $y = \log_b x$ is the mirror image of the exponential $y = b^x$ across the line $y = x$. Understanding that inverse relationship explains the entire shape — its domain, its asymptote, everything.

Logs and Exponentials Are Inverses

$y = b^x$ and $y = \log_b x$ undo each other. Feed the output of one into the other and you get back where you started:

$\log_b(b^x) = x \qquad\qquad b^{\log_b x} = x$

Because they're inverses, their graphs are reflections across the line $y = x$, and their key points swap coordinates:

💡 To plot $y = \log_b x$ fast: take any easy point on $y = b^x$ and flip its coordinates.

Features of $y = \log_b x$ (with $b > 1$)

$y = \log_b x$

The curve passes through $(1, 0)$ and $(b, 1)$, rises slowly forever to the right, and plunges toward $-\infty$ as $x$ approaches $0$.

⚠️ Domain trap: the graph never touches or crosses the $y$-axis. There is no point at $x = 0$ or for any negative $x$.

Read the Graph 🔽

Use the features of $y = \log_b x$ (base $b > 1$).

Finding the Inverse Algebraically

Because exponentials and logs are inverses, you can find one from the other by swapping $x$ and $y$.

Example: inverse of $y = 3^x$

1. Swap $x$ and $y$: $\;x = 3^y$
2. Rewrite in log form (the definition!): $\;y = \log_3 x$

So the inverse of $y = 3^x$ is $y = \log_3 x$. ✓

Example: inverse of $y = \log_5 x$

1. Swap: $\;x = \log_5 y$
2. Rewrite in exponential form: $\;y = 5^x$

🔑 Key move: converting between log and exponential form (Part 1) is the tool for finding inverses.

Concept Check 🎯

The Coordinate-Flip Shortcut

Putting the inverse idea to work: every point you know on $y = b^x$ gives you a point on $y = \log_b x$ for free — just swap the coordinates.

💡 No tables of values needed — reflecting across $y = x$ does the work. Let's try one.

Flip the Coordinates 🧮

The point $(2, 9)$ lies on $y = 3^x$ because $3^2 = 9$. Reflecting across $y = x$ gives a point on $y = \log_3 x$.

1) The reflected point on $y = \log_3 x$ has $x$-coordinate $= \,?$ 2) That same reflected point has $y$-coordinate $= \,?$ 3) Evaluate directly: $\log_3 9 = \,?$

🪵 Logarithmic Functions

Part 3 of 5 — The Laws of Logarithms

🔑 Why it matters: Logs turn multiplication into addition and powers into multiplication. These three laws are the engine behind solving equations (Part 5) and were how people multiplied huge numbers before calculators existed.

The Three Laws

For the same base $b$ (with $M, N > 0$):

Where they come from: logs are exponents, and exponents add when you multiply powers. Since $b^x \cdot b^y = b^{x+y}$, taking logs turns that product into the sum $x + y$.

⚠️ The #1 mistake: these laws apply to logs of products and quotients, not to products or sums of logs themselves. In particular: $\log_b(M + N) \ne \log_b M + \log_b N \qquad \frac{\log_b M}{\log_b N} \ne \log_b\!\left(\frac{M}{N}\right)$

Expanding a Logarithm

"Expanding" means breaking one log into a sum/difference of simpler logs.

Example: expand $\log_2(8x)$

$\log_2(8x) = \log_2 8 + \log_2 x = 3 + \log_2 x$

(Product rule, then evaluate $\log_2 8 = 3$.)

Example: expand $\log_5\!\left(\dfrac{x^3}{y}\right)$

$\log_5\!\left(\frac{x^3}{y}\right) = \log_5 x^3 - \log_5 y = 3\log_5 x - \log_5 y$

(Quotient rule first, then the power rule pulls the $3$ out front.)

💡 Order of operations, reversed: when expanding, handle the product/quotient first (outermost operation), then bring down any exponents with the power rule.

Pick the Right Law 🔽

Choose the correctly expanded form of each logarithm.

Condensing a Logarithm

"Condensing" runs the laws in reverse: combine several logs into a single log. This is the form you need before solving an equation.

Example: condense $\log_3 4 + \log_3 7$

$\log_3 4 + \log_3 7 = \log_3(4 \cdot 7) = \log_3 28$

Example: condense $2\log x - \log y$

First send coefficients back up as exponents (power rule in reverse), then combine:

$2\log x - \log y = \log x^2 - \log y = \log\!\left(\frac{x^2}{y}\right)$

🔑 Key move: to condense, coefficients must be cleared first (turn $2\log x$ into $\log x^2$) before you can merge logs with $+$ or $-$.

Concept Check 🎯

Combining Laws with Evaluation

The laws really shine when one of the pieces is a log you can evaluate exactly. The product/quotient rule splits the expression; then you collapse the known part to a number.

$\log_2(16x) = \log_2 16 + \log_2 x = 4 + \log_2 x$

$\log_2\!\left(\frac{x}{8}\right) = \log_2 x - \log_2 8 = \log_2 x - 3$

🔑 Strategy: apply a law to separate the numeric part, then replace it with its value. The next drill mixes both moves.

Apply the Laws 🧮

Use the laws, then evaluate where possible.

1) $\log_2 4 + \log_2 8 = \log_2 ? \;$ — enter the single argument (a number). 2) Evaluate that result: $\log_2 4 + \log_2 8 = \,?$ 3) $\log_b(x^5)$ expands to $?\,\log_b x$ — enter the coefficient.

🪵 Logarithmic Functions

Part 4 of 5 — Change of Base & Real-World Models

🔑 The Payoff: Calculators only have $\log$ (base $10$) and $\ln$ (base $e$) keys — yet logs appear everywhere, in earthquakes, sound, and acidity. The change-of-base formula lets you evaluate any base, and these scales show why logs matter.

The Change-of-Base Formula

To evaluate a log in any base $b$, rewrite it using a base your calculator knows:

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

You may use either common log or natural log on top and bottom — just be consistent.

Example: $\log_2 50$

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

Sanity check: $2^5 = 32$ and $2^6 = 64$, so $\log_2 50$ must be between $5$ and $6$. ✓

💡 Memory hook: "new base on the bottom." The base you're switching to (here $\ln$) goes in both spots; the original base $b$ lands in the denominator.

Concept Check 🎯

Doing It on a Calculator

The whole procedure is three keystrokes once you've set it up:

1. Write $\log_b x = \dfrac{\log x}{\log b}$ (or use $\ln$).
2. Type the numerator $\log x$, then divide by the denominator $\log b$.
3. Estimate first so you can catch an upside-down fraction.

For $\log_3 100$: since $3^4 = 81$ and $3^5 = 243$, the answer must be between $4$ and $5$ — and indeed $\frac{\log 100}{\log 3} = \frac{2}{0.477} \approx 4.19$. ✓

⚠️ Flipping the fraction is the classic error. The original base goes on the bottom — estimate to confirm you didn't invert it.

Change of Base 🧮

Use $\log_b x = \dfrac{\log x}{\log b}$. Round to two decimal places.

1) $\log_2 10 = \,?$ (use $\log 10 = 1$, $\log 2 \approx 0.301$) 2) $\log_4 64 = \,?$ (this one is exact — a whole number)

Why Logs Run on Real Scales

Several famous measurement scales are logarithmic — each whole step means a $10\times$ jump, so logs compress enormous ranges into small numbers.

Earthquake Example

A magnitude-$6$ quake is not twice a magnitude-$3$ — because each unit is a power of $10$:

$\frac{10^6}{10^3} = 10^{6-3} = 10^3 = 1000$

A magnitude-$6$ quake is $1000$ times as intense as a magnitude-$3$.

⚠️ Don't read these scales linearly. Going from pH $5$ to pH $3$ is $10^{(5-3)} = 100$ times more acidic, not "a little" more.

Logarithmic Scales 🔽

Each step on these scales is a factor of $10$.

Working the pH Formula

Chemists measure acidity with $\text{pH} = -\log[\text{H}^+]$, where $[\text{H}^+]$ is the hydrogen-ion concentration in moles per liter.

Because concentrations are tiny powers of $10$, the log makes them friendly. When $[\text{H}^+] = 10^{-n}$:

$\text{pH} = -\log(10^{-n}) = -(-n) = n$

So the pH is simply the positive version of the exponent. A neutral solution sits at pH $7$; lower is acidic, higher is basic.

💡 You'll use $\log(10^k) = k$ — one of the cleanest log facts there is.

pH Calculation 🧮

Use $\text{pH} = -\log[\text{H}^+]$, where $[\text{H}^+]$ is the hydrogen-ion concentration.

1) A solution has $[\text{H}^+] = 10^{-4}$. Its pH $= \,?$ 2) Pure water has $[\text{H}^+] = 10^{-7}$. Its pH $= \,?$

🪵 Logarithmic Functions

Part 5 of 5 — Solving Equations & Mastery Check

You can now (1) read a log as an exponent, (2) graph it as an inverse, (3) wield the three laws, and (4) change base. The final skill ties it all together: solving equations that contain logs or unknown exponents.

Solving Exponential Equations with Logs

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

Example: $2^x = 40$

$\ln(2^x) = \ln 40 \;\Rightarrow\; x\ln 2 = \ln 40 \;\Rightarrow\; x = \frac{\ln 40}{\ln 2} \approx \frac{3.689}{0.693} \approx 5.32$

Sanity check: $2^5 = 32$ and $2^6 = 64$, so $x \approx 5.32$ is reasonable. ✓

🔑 The key move: the power rule turns $\ln(2^x)$ into $x\ln 2$, freeing the exponent so you can divide for $x$.

Solving Logarithmic Equations

When the unknown is inside a log, condense to a single log, then convert to exponential form.

Example: $\log_2(x) = 5$

Convert to exponential form (the definition): $\;x = 2^5 = 32$.

Example: $\log_3(x - 1) = 2$

$x - 1 = 3^2 = 9 \;\Rightarrow\; x = 10$

⚠️ Always check for extraneous solutions. A log's argument must stay positive. Substitute back: $\log_3(10 - 1) = \log_3 9 = 2$ ✓. If a solution made any argument $\le 0$, you would reject it.

Order the Solution 🔽

You're solving $\log_2(x - 3) = 4$. Choose what happens at each stage.

Quick Reference

💡 Two identities cover most quick evaluations: $\log_b 1 = 0$ and $\log_b b = 1$.

Solve It 🧮

1) $\log_5(x) = 3 \;\Rightarrow\; x = \,?$ 2) $\log_2(x + 5) = 4 \;\Rightarrow\; x = \,?$ 3) $10^x = 1000 \;\Rightarrow\; x = \,?$

Two Traps to Avoid

Before the mixed set, lock in the two mistakes that sink the most students:

⚠️ Trap 1 — dropping the check. After solving a log equation, substitute back and confirm every argument is positive. Reject any root that makes a log's argument $\le 0$.

⚠️ Trap 2 — confusing the two equation types. If the unknown is in the exponent ($b^x = k$), take a log. If the unknown is inside the log ($\log_b(\cdot) = c$), convert to exponential form. Pick the move that frees the variable.

Mixed Practice 🎯

You've Mastered Logarithms

From a single idea — a log is an exponent — you built the whole toolkit:

• Read & switch forms: $\log_b x = y \Leftrightarrow b^y = x$
• Graph: the inverse of $b^x$, domain $x > 0$, asymptote at $x = 0$
• Laws: product → sum, quotient → difference, power → coefficient
• Change of base and real-world scales (Richter, decibels, pH)
• Solve exponential and logarithmic equations (and check for extraneous roots)

🔑 One last time: when in doubt, rewrite the log as an exponent. That move unlocks almost everything. Now finish strong on the Exit Quiz.

Exit Quiz ✅

Answer all three to finish the lesson.