9709 P3 §3.2 — Logarithms & Indices

💡 Lead In

The Scenario

Imagine a single bacterium in a petri dish. Every hour, each bacterium splits into two. After 1 hour: 2 bacteria. After 2 hours: 4. After 3 hours: 8. The population grows explosively — after just 20 hours, there would be over a million bacteria from that single starting cell.

The Question

You know the pattern: population $= 2^t$ after $t$ hours. So you need to solve $2^t = 1\,000\,000$. You could guess and check — $2^{10} = 1024$, $2^{20} = 1\,048\,576$ — but what if the number weren't a power of 2 at all? Is there a mathematical operation that "undoes" exponentiation, the way division undoes multiplication?

Quick Review: Exponential Laws

Logarithms are built on the index (exponential) laws. Before we introduce $\log$, let's make sure the three exponential laws are fresh. Select the correct answer for each.

Q1 (Product law) — $a^m \times a^n = ?$

$a^{mn}$

$a^{m+n}$

$a^{m-n}$

$a^{n-m}$

Product law: when multiplying powers of the same base, add the exponents. Example: $2^3 \times 2^5 = 2^{3+5} = 2^8$.

Q2 (Quotient law) — $a^m \div a^n = ?$

$a^{m-n}$

$a^{m+n}$

$a^{mn}$

$a^{m/n}$

Quotient law: when dividing powers of the same base, subtract the exponents. Example: $\dfrac{3^7}{3^2} = 3^{7-2} = 3^5$.

Q3 (Power law) — $(a^m)^n = ?$

$a^{m+n}$

$a^{m-n}$

$a^{mn}$

$a^{m^n}$

Power law: raising a power to a power multiplies the exponents. Example: $(5^2)^3 = 5^{2 \times 3} = 5^6$.

📖 Key Concepts

Definition: The logarithm of $b$ to base $a$ is the power to which $a$ must be raised to give $b$. In symbols: $$a^x = b \iff \log_a(b) = x \quad (a > 0,\ a \neq 1,\ b > 0)$$ A logarithm asks a question: "$a$ to what power equals $b$?"

Explore: adjust the base $a$ and the exponent $x$. Watch how the exponential form and logarithmic form stay linked.

a: 3

x: 2

Exponential form: $3^{2} = 9$ Logarithmic form: $\log_{3}(9) = 2$

Your mission: In the activities below, you will derive the three laws of logarithms algebraically -- the Product Law, Quotient Law, and Power Law. Each derivation follows the same strategy: convert to exponential form, apply an index law, then convert back. You will complete each step yourself, choosing the correct algebraic expression at every stage.

Activity 1 · Pair Log Law Escape Room — Derive the Product & Quotient Laws

Think-First (60 s silent)

Recall the definition $a^x = b \iff \log_a(b) = x$ and the index laws $a^m \times a^n = a^{m+n}$, $\dfrac{a^m}{a^n} = a^{m-n}$. With your partner, silently think: If $\log_a(m) = x$ means $m = a^x$, can you express $mn$ and $\dfrac{m}{n}$ in terms of $a$, $x$, and $y$? The escape room below will guide you through the full algebraic derivation, step by step.

Escape Room rules (8-10 min): Work in pairs. Two locked doors stand between you and freedom. Each door requires you to complete a 4-step algebraic derivation. At each step, choose the correct expression from the options. Correct on the first try earns full points. Unlock both doors to escape!

🔐 Log Law Laboratory — Escape Room Doors Unlocked: 0 / 2 | Score: 0 / 40

🔒 Door 1 · Product Law Step 1 / 4 | Score: 0 / 20

Product Law — Proved!

$$\log_a(mn) = \log_a(m) + \log_a(n)$$

🔒 Door 2 · Quotient Law Step 1 / 4 | Score: 0 / 20

Quotient Law — Proved!

$$\log_a\!\left(\frac{m}{n}\right) = \log_a(m) - \log_a(n)$$

🏆

Both Doors Unlocked — You Escaped!

You have algebraically derived the Product Law and the Quotient Law. Total score: 0 / 40

Activity 2 · Individual Power Law Slow Reveal — Predict Each Derivation Step

Think-First (30 s silent)

You have derived the Product and Quotient Laws. Now for the Power Law: $\log_a(m^n) = n \cdot \log_a(m)$. Silently think: What index law connects $(a^x)^n$ to a single power of $a$? In this activity you will predict each step of the algebraic derivation before it is revealed.

Slow Reveal rules (5-6 min): Work individually. The derivation is revealed one step at a time. Before each step appears, predict what comes next by choosing from the options. Correct predictions on the first try earn full points. Complete all 5 steps to reveal the full derivation chain!

🔍 Power Law — Slow Reveal Derivation Step 3 / 5 | Score: 12 / 20 | ★☆☆

1 Setup — Substitution

We want to prove $\log_a(m^n) = n \cdot \log_a(m)$. First, set up by letting $\log_a(m) = x$. What does this give us by the definition of a logarithm?

👇 Predict the next step:

Correct! 4 points.

Let $\log_a(m) = x$, so by the definition of a logarithm: $a^x = m$.

2 Raise to Power $n$

We have $m = a^x$. We need $m^n$ on the left side. What should we do to both sides?

👇 Predict the next step:

Correct! 4 points.

Raise both sides to the power $n$: $m^n = (a^x)^n$.

3 Apply Index Law

We have $m^n = (a^x)^n$. Apply the index law $(a^x)^n = a^{xn}$ to simplify the right side.

👇 Predict the simplified form:

Correct! 4 points.

By the index law $(a^p)^q = a^{pq}$: $m^n = a^{xn}$.

4 Convert to Base-$a$ Logarithms

We have $m^n = a^{xn}$. By the definition of a logarithm ($a^p = b \iff \log_a(b) = p$), convert the right-hand side to logarithmic form. What is $\log_a(m^n)$?

👇 Predict the result:

5 Substitute Back — Final Law

We have $\log_a(m^n) = xn$. Substitute back $x = \log_a(m)$ to obtain the Power Law.

👇 Predict the final law:

🔗 Complete Derivation Chain

1 Let $\log_a(m) = x$, so by definition $a^x = m$.

2 Raise both sides to power $n$: $m^n = (a^x)^n$.

3 Apply index law: $m^n = a^{xn}$.

4 Convert the right-hand side to logarithmic form: using the definition $a^{xn} = m^n \iff \log_a(m^n) = xn$.

5 Substitute $x = \log_a(m)$: $\log_a(m^n) = n \cdot \log_a(m)$.

Power Law — Proved!

$$\log_a(m^n) = n\log_a(m)$$

🏆

Power Law Derived!

You have predicted every step of the derivation. Total score: 0 / 20

Common Logarithm Introducing Common Logarithms — Base 10

Definition: When a logarithm is written without a base, e.g. $\log x$, the base is understood to be 10. This is called the common logarithm.

$$\log x = \log_{10} x$$

All three laws you derived apply to common logarithms in exactly the same way:

Product: $\log(mn) = \log m + \log n$
Quotient: $\log\!\left(\frac{m}{n}\right) = \log m - \log n$
Power: $\log(m^n) = n\log m$

Activity 3 · Individual Rapid-Fire Log Quiz — Speed & Accuracy Challenge

Rapid-Fire rules: Answer each question before the timer runs out! Questions appear one at a time — choose the correct answer for instant feedback. Green = correct, Red = wrong. Your final score depends on both accuracy and remaining time. Good luck!

20 questions across three types:

✅ True/False 🔢 Calculation 📋 Apply the Law

Timer: 100 seconds | Scoring: +10 per correct + time bonus

✏️ Worked Examples

Worked Example 1 · Group Writing Expressions in the Form log x

Ex 2B Q4 Standard

Write the following expression in the form $\log x$ where $x$ is a number.

(x) $2\log 4 + \log 9 - \dfrac{1}{2}\log 144$

▼ Reveal Solution

Part (x): Combined

$2\log 4 + \log 9 - \dfrac{1}{2}\log 144$

$= \log(4^2) + \log 9 - \log(144^{1/2})$ (power law)

$= \log 16 + \log 9 - \log 12$

$= \log\!\left(\dfrac{16 \times 9}{12}\right)$ (product + quotient law)

$= \log\!\left(\dfrac{144}{12}\right) = \log 12$

Final Answer

(x) $\log 12$

Worked Example 2 · Group Fractional Exponents in Terms of log x

Ex 2B Q5(iv) Standard

Express the following in terms of $\log x$.

(iv) $\log x^{\frac{3}{2}} + \log \sqrt[3]{x}$

▼ Reveal Solution

Part (iv): Fractional exponents

$\log x^{\frac{3}{2}} + \log \sqrt[3]{x}$

$= \log x^{\frac{3}{2}} + \log x^{\frac{1}{3}}$ (write $\sqrt[3]{x}$ as $x^{1/3}$)

$= \dfrac{3}{2}\log x + \dfrac{1}{3}\log x$ (power law)

$= \left(\dfrac{9}{6} + \dfrac{2}{6}\right)\log x = \dfrac{11}{6}\log x$ (collect like terms)

Final Answer

(iv) $\dfrac{11}{6}\log x$

🎯 Concept Quiz — Quick Check

Test your understanding of all four lesson objectives. Each question targets one objective.

Score: 0 / 4

Q1 (Definition) — If $5^x = 125$, which expression equals $x$?

$x = \log_{125} 5$

$x = \log_5 125$

$x = \dfrac{125}{5}$

$x = \log 125$

By definition $a^x = b \iff \log_a(b) = x$. Here $5^x = 125$, so $x = \log_5 125$. (And since $5^3 = 125$, we have $x = 3$.)

Q2 (Product law) — Simplify $\log_3 6 + \log_3 5$ as a single logarithm:

$\log_3 11$

$\log_3 1$

$\log_3\!\left(\dfrac{6}{5}\right)$

$\log_3 30$

Product law: $\log_a(mn) = \log_a(m) + \log_a(n)$. So $\log_3 6 + \log_3 5 = \log_3(6 \times 5) = \log_3 30$.

Q3 (Quotient law) — Simplify $\log_2 48 - \log_2 6$ as a single logarithm:

$\log_2 8$

$\log_2 42$

$\log_2 54$

$\log_2 288$

Quotient law: $\log_a\!\left(\dfrac{m}{n}\right) = \log_a(m) - \log_a(n)$. So $\log_2 48 - \log_2 6 = \log_2\!\left(\dfrac{48}{6}\right) = \log_2 8$. (And since $2^3 = 8$, this equals $3$.)

Q4 (Power law) — Simplify $3\log_5 4$ as a single logarithm:

$\log_5 12$

$\log_5 7$

$\log_5 64$

$\log_5\!\left(\dfrac{4}{3}\right)$

Power law: $n\,\log_a(m) = \log_a(m^n)$. So $3\log_5 4 = \log_5(4^3) = \log_5 64$.

🎉

You scored 0 / 4

✏️ More Practice

Exercise 2B Q1 Standard

$2^x = 32 \iff x = \log_2 32$. Write similar logarithmic equivalents of these equations. In each case find also the value of $x$, using your knowledge of indices and not using your calculator.

(i) $3^x = 9$ (ii) $4^x = 64$ (iii) $2^x = \dfrac{1}{4}$ (iv) $5^x = \dfrac{1}{5}$ (v) $7^x = 1$ (vi) $16^x = 2$

▼ Show Solution

Solution

(i) $3^x = 9 \iff x = \log_3 9$, and $x = 2$ (since $3^2 = 9$).

(ii) $4^x = 64 \iff x = \log_4 64$, and $x = 3$ (since $4^3 = 64$).

(iii) $2^x = \dfrac{1}{4} \iff x = \log_2\!\left(\dfrac{1}{4}\right)$, and $x = -2$ (since $2^{-2} = \dfrac{1}{4}$).

(iv) $5^x = \dfrac{1}{5} \iff x = \log_5\!\left(\dfrac{1}{5}\right)$, and $x = -1$ (since $5^{-1} = \dfrac{1}{5}$).

(v) $7^x = 1 \iff x = \log_7 1$, and $x = 0$ (since $7^0 = 1$).

(vi) $16^x = 2 \iff x = \log_{16} 2$, and $x = \dfrac{1}{4}$ (since $16^{1/4} = 2$).

Exercise 2B Q2 Standard

Write the equivalent of these equations in exponential form. Without using your calculator, find also the value of $y$ in each case.

(i) $y = \log_3 9$ (ii) $y = \log_5 125$ (iii) $y = \log_2 16$ (iv) $y = \log_6 1$ (v) $y = \log_{64} 8$ (vi) $y = \log_5\!\left(\dfrac{1}{25}\right)$

▼ Show Solution

Solution

(i) $y = \log_3 9 \iff 3^y = 9$, so $y = 2$.

(ii) $y = \log_5 125 \iff 5^y = 125$, so $y = 3$.

(iii) $y = \log_2 16 \iff 2^y = 16$, so $y = 4$.

(iv) $y = \log_6 1 \iff 6^y = 1$, so $y = 0$.

(v) $y = \log_{64} 8 \iff 64^y = 8$, so $y = \dfrac{1}{2}$ (since $64^{1/2} = 8$).

(vi) $y = \log_5\!\left(\dfrac{1}{25}\right) \iff 5^y = \dfrac{1}{25}$, so $y = -2$.

Exercise 2B Q3 Medium

Write down the values of the following without using a calculator. Use your calculator to check your answers for those questions which use base 10.

(i) $\log_{10} 10000$ (ii) $\log_{10}\!\left(\dfrac{1}{10000}\right)$ (iii) $\log_{10} \sqrt{10}$ (iv) $\log_{10} 1$ (v) $\log_{3} 81$ (vi) $\log_{3}\!\left(\dfrac{1}{81}\right)$ (vii) $\log_{3} \sqrt{27}$ (viii) $\log_{3} \sqrt[4]{3}$ (ix) $\log_{4} 2$ (x) $\log_{5}\!\left(\dfrac{1}{125}\right)$

▼ Show Solution

Solution

(i) $\log_{10} 10000 = 4$ (since $10^4 = 10000$)

(ii) $\log_{10}\!\left(\dfrac{1}{10000}\right) = -4$ (since $10^{-4} = \dfrac{1}{10000}$)

(iii) $\log_{10} \sqrt{10} = \dfrac{1}{2}$ (since $10^{1/2} = \sqrt{10}$)

(iv) $\log_{10} 1 = 0$ (since $10^0 = 1$)

(v) $\log_{3} 81 = 4$ (since $3^4 = 81$)

(vi) $\log_{3}\!\left(\dfrac{1}{81}\right) = -4$ (since $3^{-4} = \dfrac{1}{81}$)

(vii) $\log_{3} \sqrt{27} = \dfrac{3}{2}$ (since $\sqrt{27} = 27^{1/2} = (3^3)^{1/2} = 3^{3/2}$)

(viii) $\log_{3} \sqrt[4]{3} = \dfrac{1}{4}$ (since $\sqrt[4]{3} = 3^{1/4}$)

(ix) $\log_{4} 2 = \dfrac{1}{2}$ (since $4^{1/2} = 2$)

(x) $\log_{5}\!\left(\dfrac{1}{125}\right) = -3$ (since $5^{-3} = \dfrac{1}{125}$)

Exercise 2B Q8 Medium

Express the following as a single logarithm. $2\log_{10}x - \log_{10}7$

Hence solve $2\log_{10}x - \log_{10}7 = \log_{10}63$

💡 Hint

Hint

First use the power law to write $2\log_{10}x = \log_{10}(x^2)$. Then combine using the quotient law. For the equation, if $\log_{10} A = \log_{10} B$, then $A = B$ (provided $A, B > 0$).

▼ Show Solution

Step 1: Express as a single logarithm

Using the power law: $2\log_{10}x = \log_{10}(x^2)$.

Using the quotient law:

$$2\log_{10}x - \log_{10}7 = \log_{10}(x^2) - \log_{10}7 = \log_{10}\!\left(\dfrac{x^2}{7}\right)$$

Step 2: Solve the equation

The equation becomes:

$$\log_{10}\!\left(\dfrac{x^2}{7}\right) = \log_{10}63$$

Since the logarithm function is one-to-one, equate the arguments:

$$\dfrac{x^2}{7} = 63 \implies x^2 = 441 \implies x = \pm 21$$

Step 3: Check validity

Since $\log_{10}x$ requires $x > 0$, we reject $x = -21$.

$$x = 21$$

Check

$2\log_{10}21 - \log_{10}7 = \log_{10}(21^2) - \log_{10}7 = \log_{10}(441) - \log_{10}7 = \log_{10}\!\left(\dfrac{441}{7}\right) = \log_{10}63$. ✓

🎫

Exit Ticket

Check the learning objectives you feel confident about:

I can convert between exponential and logarithmic forms using $a^x = b \iff \log_a(b) = x$. I can apply the product law: $\log_a(mn) = \log_a(m) + \log_a(n)$. I can apply the quotient law: $\log_a\!\left(\dfrac{m}{n}\right) = \log_a(m) - \log_a(n)$. I can apply the power law: $\log_a(m^n) = n\,\log_a(m)$ to simplify and combine logarithmic expressions.