Module 13 of 47Lesson 1 of 7 · HL · AI 1.9 — Laws of logarithms

Course map · Topic 1: Number and Algebra

1.9 Laws of logarithms

HL only11 worked questions

The three laws of logarithms turn multiplication into addition, division into subtraction and powers into multiplication. They let you solve exponential equations by hand, combine or split log expressions, and they are the reason log scales and linearised graphs work.

✏️ Start with the questions. Try each one on paper, then open the solution and compare.

Stuck? Open the notes ↓

✏️ Practice questions

Try each question on paper first. Then open the solution and compare every step.

Question 1●○○ Warm-up✍️ By hand · Paper 1[3 marks]

Write each expression as a single logarithm.

(a) $\log 5 + \log 4$ [1]
(b) $\ln 18 - \ln 6$ [1]
(c) $3\log 2 + \log 5$ [1]

Show the full step-by-step solution

Step 1🧠 ThinkRead the laws backwards

Plus outside → multiply inside. Minus outside → divide inside. A number in front → becomes a power inside. When there is a number in front, move it up as a power first.

Part (a)

Step 1✍️ By handAdd → multiply

\log 5 + \log 4 = \log(5\times4) = \log 20

Part (b)

Step 1✍️ By handSubtract → divide

\ln 18 - \ln 6 = \ln\left(\dfrac{18}{6}\right) = \ln 3

Part (c)

Step 1✍️ By handPower first, then add

3\log 2 = \log 2^3 = \log 8

\log 8 + \log 5 = \log 40

Step 2🖩 On the GDCCheck (a) and (b) with numbers

Same values, so the laws worked

Final answer (a)

\log 20

(b)

\ln 3

(c)

\log 40

Question 2●○○ Warm-up✍️ By hand · Paper 1[3 marks]

Given that

x, y, z \gt 0

, write

\log\left(\dfrac{x^2y}{z}\right)

in terms of

\log x

\log y

and

\log z

Show the full step-by-step solution

Step 1🧠 ThinkSplit, then bring powers down

Inside the log,

x^2

and

y

are multiplied (top) and

z

divides (bottom). Top pieces get a plus, bottom pieces get a minus. Then the power 2 comes down to the front.

Step 2✍️ By handUse the quotient and product laws

\log\left(\dfrac{x^2y}{z}\right) = \log x^2 + \log y - \log z

Step 3✍️ By handUse the power law

= 2\log x + \log y - \log z

A1A1

Final answer

2\log x + \log y - \log z

Question 3●○○ Warm-up✍️ By hand · Paper 1[4 marks]

Let

\log 2 = p

and

\log 3 = q

. Write each of the following in terms of

p

and

q

(a) $\log 12$ [2]
(b) $\log 1.5$ [1]
(c) $\log 0.8$ [1]

Show the full step-by-step solution

Step 1🧠 ThinkBuild each number out of 2s, 3s and 10s

We only know

\log 2

\log 3

and also

\log 10 = 1

(because

10^1 = 10

). So write each number using only 2, 3 and 10 multiplied or divided.

Part (a)

Step 1✍️ By hand12 = 2² × 3

\log 12 = \log(2^2\times3) = 2\log 2 + \log 3

= 2p + q

Part (b)

Step 1✍️ By hand1.5 = 3 ÷ 2

\log 1.5 = \log\dfrac{3}{2} = q - p

Part (c)

Step 1✍️ By hand0.8 = 8 ÷ 10 = 2³ ÷ 10

\log 0.8 = \log\dfrac{2^3}{10} = 3\log2 - \log10 = 3p - 1

✅ Check it With

p = 0.30103

and

q = 0.47712

2p+q = 1.07918 = \log 12

✔ and

3p - 1 = -0.09691 = \log 0.8

✔

Final answer (a)

2p + q

(b)

q - p

(c)

3p - 1

Question 4●●○ Standard✍️ + 🖩 · Paper 1[4 marks]

Logarithms with bases other than 10 and

e

(a) Solve $3^x = 40$ , giving your answer to 3 significant figures.[2]
(b) Find the value of $\log_7 50$ .[2]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkTake logs of both sides

If two things are equal, their logs are equal. Taking

\ln

(or

\log

) of both sides lets the power law bring

x

down to the front.

Step 2✍️ By handPower law

\ln 3^x = \ln 40

x\ln 3 = \ln 40

x = \dfrac{\ln 40}{\ln 3} = 3.35776\ldots \approx 3.36

Part (b)

Step 1✍️ By handChange of base

\log_7 50 = \dfrac{\ln 50}{\ln 7}

= 2.01038\ldots \approx 2.01

Step 2🖩 On the GDCOn the GDC: log with a base

log(40,3)›enter💻 Enter›log(50,7)›enter💻 Enter

📍 Where to find it on the TI-Nspire

⌨️ log: ctrl then 10ˣ on the keypad, or type log( (for base b type log(x,b))

Type log(40,3) — the second number is the base. It is shown as

\log_3(40)

. You can also use the

\log_\square

template from the templates key.

Both routes agree

✅ Check it

3^3 = 27

and

3^4 = 81

, so

x

must be between 3 and 4. ✔ And

7^2 = 49

, so

\log_7 50

is just above 2. ✔

Final answer (a)

x \approx 3.36

(b)

2.01

Question 5●●○ Standard✍️ + 🖩 · Paper 1[4 marks]

Solve the equation

5^{2x-1} = 200

. Give your answer correct to 3 significant figures.

Show the full step-by-step solution

Step 1🧠 ThinkThe whole bracket is the power

Take

\ln

of both sides. The power law brings the whole power

(2x-1)

down. Then it is an ordinary linear equation.

Step 2✍️ By handTake logs and solve

\ln 5^{2x-1} = \ln 200

(2x-1)\ln 5 = \ln 200

2x - 1 = \dfrac{\ln 200}{\ln 5} = 3.29203\ldots

2x = 4.29203\ldots

x = 2.14601\ldots \approx 2.15

Step 3🖩 On the GDCCheck with nSolve

menu💻 click menu on keypad›3: Algebra›1: Numerical Solve

x = 2.14601

Final answer

x \approx 2.15

⚠️ Common mistake Writing

2x - 1\ln5

without a bracket, then only multiplying the 1 by

\ln 5

Question 6●●○ Standard✍️ By hand · Paper 1[5 marks]

Solve

\log_2 x + \log_2(x+2) = 3

Show the full step-by-step solution

Step 1🧠 ThinkMake one log, then remove it

Two logs added together become one log of a product. Then “

\log_2(\text{something}) = 3

” means “something

= 2^3

”. That gives a normal quadratic.

Step 2✍️ By handProduct law

\log_2\big(x(x+2)\big) = 3

Step 3✍️ By handChange to exponential form

x(x+2) = 2^3 = 8

x^2 + 2x - 8 = 0

(x+4)(x-2) = 0 \Rightarrow x = -4 \text{ or } x = 2

Step 4✍️ By handCheck which answers are allowed

We cannot take

\log_2

of a negative number, so

x = -4

does not work.

Reject

x = -4

because

\log_2(-4)

is undefinedR1

x = 2

✅ Check it

\log_2 2 + \log_2 4 = 1 + 2 = 3

✔

Final answer

x = 2

⚠️ Common mistake Giving both

x = -4

and

x = 2

. Always check the solutions in the original logs.

Question 7●●○ Standard✍️ + 🖩 · Paper 1[6 marks]

Solve

2\ln x - \ln(x+6) = \ln 2

, for

x \gt 0

. Give your answer in the form

a + \sqrt{b}

, where

a, b \in \mathbb{Z}

Show the full step-by-step solution

Step 1🧠 ThinkCollect everything into one log

First move the 2 up as a power:

2\ln x = \ln x^2

. Then the minus becomes a division. When

\ln(\text{left}) = \ln(\text{right})

, the insides must be equal.

Step 2✍️ By handPower law, then quotient law

\ln x^2 - \ln(x+6) = \ln 2

\ln\left(\dfrac{x^2}{x+6}\right) = \ln 2

\dfrac{x^2}{x+6} = 2

Step 3✍️ By handSolve the quadratic

x^2 = 2x + 12 \Rightarrow x^2 - 2x - 12 = 0

x = \dfrac{2 \pm \sqrt{4 + 48}}{2} = \dfrac{2\pm\sqrt{52}}{2} = 1 \pm \sqrt{13}

1 - \sqrt{13} \approx -2.61 \lt 0

: rejectR1

x = 1 + \sqrt{13}

Step 4🖩 On the GDCCheck with nSolve

menu💻 click menu on keypad›3: Algebra›1: Numerical Solve

1 + √13 = 4.60555

Final answer

x = 1 + \sqrt{13}\ (\approx 4.61)

Question 8●●○ Standard✍️ + 🖩 · Paper 2[5 marks]

Solve

4^{x} = 3^{x+2}

. Give your answer correct to 3 significant figures.

Show the full step-by-step solution

Step 1🧠 ThinkDifferent bases — take ln of both sides

The bases 4 and 3 are not powers of the same number. Take

\ln

of both sides, bring the powers down, then collect all the

x

terms on one side.

Step 2✍️ By handBring the powers down

x\ln 4 = (x+2)\ln 3

x\ln 4 = x\ln 3 + 2\ln 3

x(\ln 4 - \ln 3) = 2\ln 3

x = \dfrac{2\ln 3}{\ln\frac{4}{3}}

x = 7.63768\ldots \approx 7.64

Step 3🖩 On the GDCCheck on a graph: the two sides after taking ln

menu💻 click menu on keypad›6: Analyze Graph›4: Intersection

Graphing

4^x

and

3^{x+2}

needs a huge y-window (they meet at about 39 659). It is neater to graph the two sides after taking ln: f1(x) = x·ln(4) and f2(x) = (x+2)·ln(3). They are straight lines with the same intersection x-value.

Final answer

x = \dfrac{2\ln 3}{\ln 4 - \ln 3} \approx 7.64

Question 9●●● Exam-level✍️ + 🖩 · Paper 2[7 marks]

Neo invests P 5 000 in account A, which grows by 6% each year. Onalenna invests P 4 000 in account B, which grows by 8% each year. After

t

years the values (in pula) are

A = 5000\times1.06^{t}

and

B = 4000\times1.08^{t}

(a) Show that the accounts have equal value when $t = \dfrac{\ln 1.25}{\ln\left(\frac{1.08}{1.06}\right)}$ .[3]
(b) Hence find $t$ and the value of each account at that time.[2]
(c) Find the number of complete years after which account B is first worth more than account A.[2]

Show the full step-by-step solution

Part (a)

Step 1🧠 Think“Show that” — every step must be written

Set

A = B

. Put the powers together on one side and the numbers on the other. Then take

\ln

. The final line must match the given expression exactly.

Step 2✍️ By handSet equal and rearrange

5000\times1.06^{t} = 4000\times1.08^{t}

\dfrac{1.08^{t}}{1.06^{t}} = \dfrac{5000}{4000} = 1.25

\left(\dfrac{1.08}{1.06}\right)^{t} = 1.25

t\ln\left(\dfrac{1.08}{1.06}\right) = \ln 1.25

t = \dfrac{\ln 1.25}{\ln\left(\frac{1.08}{1.06}\right)}

Part (b)

Step 1✍️ By handEvaluate

t = 11.9378\ldots \approx 11.9

yearsA1

A = B = 5000\times1.06^{11.9378\ldots} \approx \text{P }10\,024.60

Step 2🖩 On the GDCGraph both accounts and find the intersection

menu💻 click menu on keypad›6: Analyze Graph›4: Intersection

Part (c)

Step 1✍️ By handRound up and check both sides of 11.94

t = 11

A = 9491.49,\ B = 9326.56

→ A still biggerM1

t = 12

A = 10\,060.98,\ B = 10\,072.68

→ B bigger. Answer: 12 yearsA1

Final answer (a) shown (b)

t \approx 11.9

years, each worth about

\text{P }10\,024.60

Question 10●●● Exam-level✍️ + 🖩 · Paper 2[7 marks]

Two variables are thought to follow a power model

y = kx^{n}

, where

k, n \gt 0

(a) Show that $\log y = n\log x + \log k$ .[2]
(b) A graph of $\log y$ against $\log x$ is a straight line with gradient $1.5$ and vertical intercept $0.301$ . Find $n$ and $k$ .[3]
(c) Hence estimate $y$ when $x = 16$ .[2]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkTake log of both sides, then split

The right side is a product,

k\times x^n

, so the product law splits it. Then the power law brings

n

down.

Step 2✍️ By handProduct law, then power law

\log y = \log(kx^{n}) = \log k + \log x^{n}

\log y = n\log x + \log k

Part (b)

Step 1🧠 ThinkCompare with Y = mX + c

If we call

Y = \log y

and

X = \log x

, the equation is

Y = nX + \log k

: a straight line with gradient

n

and intercept

\log k

Step 2✍️ By handRead off n, then undo the log for k

n = 1.5

\log k = 0.301 \Rightarrow k = 10^{0.301}

k = 1.99986\ldots \approx 2.00

Part (c)

Step 1✍️ By handSubstitute into the model

y = 10^{0.301}\times16^{1.5}

y = 127.991\ldots \approx 128

Step 2🖩 On the GDCCalculator values

k and the estimate

Final answer (a) shown (b)

n = 1.5

k \approx 2.00

(c)

y \approx 128

⚠️ Common mistake Taking

k = 0.301

. The intercept is

\log k

, not

k

— you must undo the log with

10^{x}

Question 11●●● Exam-level✍️ + 🖩 · Paper 2[6 marks]

The magnitude

M

of an earthquake on a seismograph is

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

, where

A

is the amplitude of the ground shaking and

A_0

is a fixed reference amplitude.

(a) Two earthquakes have magnitudes $M_1$ and $M_2$ , and amplitudes $A_1$ and $A_2$ . Show that $M_1 - M_2 = \log\left(\dfrac{A_1}{A_2}\right)$ .[2]
(b) An earthquake felt in southern Africa had magnitude 6.5. A small tremor near Francistown had magnitude 4.2. Find how many times larger the amplitude of the first earthquake was.[2]
(c) Find the increase in magnitude when the amplitude is multiplied by 50.[2]

Show the full step-by-step solution

Part (a)

Step 1✍️ By handQuotient law on each, then subtract

M_1 - M_2 = (\log A_1 - \log A_0) - (\log A_2 - \log A_0)

= \log A_1 - \log A_2

= \log\left(\dfrac{A_1}{A_2}\right)

Part (b)

Step 1🧠 ThinkUse part (a)

The difference in magnitude is

6.5 - 4.2 = 2.3

. That equals the log of the amplitude ratio, so undo the log with

10^{x}

Step 2✍️ By handUndo the log

\log\left(\dfrac{A_1}{A_2}\right) = 2.3 \Rightarrow \dfrac{A_1}{A_2} = 10^{2.3}

= 199.526\ldots \approx 200

timesA1

Part (c)

Step 1✍️ By handRatio 50 → difference log 50

M_1 - M_2 = \log 50

= 1.69897\ldots \approx 1.70

Step 2🖩 On the GDCCalculator values

(b) and (c)

Final answer (a) shown (b) about

200

times (

199.526\ldots

) (c) an increase of

1.70

⚠️ Common mistake Thinking a difference of 2.3 in magnitude means 2.3 times bigger. Each 1 unit of magnitude multiplies the amplitude by 10.

🆘 Stuck on a question? Open the notes, GDC toolkit and common mistakes

💻 Using the TI-Nspire CX Student Software on a laptop

The screens in these notes are drawn exactly as the Student Software shows them. The keypad on the left of the software works like the handheld — click its keys with the mouse.
menu: click menu on the on-screen keypad (or Documents Toolbox ▸ 🔧). The menu numbers and letters are the same as in these notes.
enter = Enter · tab = Tab (next box) · esc = Esc · del = Backspace.
Powers: type ^ (Shift+6), the power, then press → to come back down before typing more. x² = ^2 then →.
Fraction: type / · √: type sqrt( · π: type pi · eˣ: type exp( · store: type :=. For EE (×10ⁿ) and i, click them on the keypad.
New page: Ctrl+I (or the Insert menu) · change page: Ctrl+← / Ctrl+→ · graph entry line: Ctrl+G · var: Ctrl+L.
Radian / Degree: click Settings in the status bar at the bottom of the window. The mode (RAD or DEG) shows in the top bar.
IB exam mode (Press-to-Test): everything in these notes works on the non-CAS TI-Nspire with the IB exam settings. In exam mode there are no dot-product, cross-product, unit-vector or eigenvalue keys and no Diff Eq graphs — the notes show the allowed way (type the components, use polyRoots on the characteristic equation, do Euler steps on a Calculator page).
If you type nDeriv(, the software shows it as centralDiff( — it is the same command.

🎯 What you must know

Logs are powers: $\log_a x$ is the power you put on $a$ to get $x$ . So $\log_a a = 1$ and $\log_a 1 = 0$ .
Product law: $\log_a(xy) = \log_a x + \log_a y$ — multiply inside → add outside.
Quotient law: $\log_a\left(\dfrac{x}{y}\right) = \log_a x - \log_a y$ — divide inside → subtract outside.
Power law: $\log_a x^m = m\log_a x$ — a power inside comes down to the front. This is how you get an unknown out of the power.
Change of base: $\log_a x = \dfrac{\log x}{\log a} = \dfrac{\ln x}{\ln a}$ . The GDC also accepts $\log(x, a)$ for $\log_a x$ .
The laws only work for positive $x$ and $y$ . After solving a log equation, reject any answer that would make a log of zero or a negative number.
There is no law for $\log(x+y)$ : it is not $\log x + \log y$ .

Laws of logarithms

\log_a xy = \log_a x + \log_a y,\quad \log_a\dfrac{x}{y} = \log_a x - \log_a y,\quad \log_a x^m = m\log_a x

📘 In the formula booklet

Exponents and logarithms

a^x = b \iff x = \log_a b

📘 In the formula booklet

Change of base

\log_a x = \dfrac{\ln x}{\ln a} = \dfrac{\log x}{\log a}

🧠 Learn this one

🖩 Your GDC toolkit for this subtopic

Follow the steps on your own TI-Nspire CX II. The screens show what you should see.

Log to any base: type log(value, base), or use the log template

templates💻 click ⊞ on keypad›log(40,3)›enter💻 Enter

📍 Where to find it on the TI-Nspire

⌨️ log: ctrl then 10ˣ on the keypad, or type log( (for base b type log(x,b))

Press the templates key and pick the template.
Choose log(40,3).
Press enter.

Both routes agree

Worked example: Question 4 — On the GDC: log with a base

Type log(40,3) — the second number is the base. It is shown as

\log_3(40)

. You can also use the

\log_\square

template from the templates key.

Solve an exponential or log equation

menu💻 click menu on keypad›3: Algebra›1: Numerical Solve

Start on a Calculator page (ctrl doc adds one).
Press menu.
Choose 3: Algebra.
Choose 1: Numerical Solve.

The menu path

x = 2.14601

Worked example: Question 5 — Check with nSolve

Graph both sides and find where they meet

menu💻 click menu on keypad›6: Analyze Graph›4: Intersection

Start on a Graphs page (ctrl doc adds one).
Press menu.
Choose 6: Analyze Graph.
Choose 4: Intersection.
Move the pointer to the LEFT of the crossing point and press click, then to the RIGHT and press click. The intersection point and its coordinates appear.

The menu path

Worked example: Question 8 — Check on a graph: the two sides after taking ln

Graphing

4^x

and

3^{x+2}

⚠️ Where students lose marks

Writing $\log(x+2) = \log x + \log 2$ . The product law needs a product inside the log, not a sum.
Writing $\dfrac{\log 40}{\log 3} = \log\dfrac{40}{3}$ . A fraction of two logs is not the log of a fraction.
Forgetting to check answers: $x = -4$ cannot go into $\log_2 x$ , so it must be rejected (R1 mark).
Bringing down only part of a power: $\ln 5^{2x-1} = (2x-1)\ln 5$ — the whole bracket comes down.

IB DP Mathematics: Applications and Interpretation (AI) — SL and HL HL · AI 1.9 — Laws of logarithms

IB DP Mathematics: Applications and Interpretation (AI) — SL and HL

1.9 Laws of logarithms

✏️ Practice questions

🎯 What you must know

🖩 Your GDC toolkit for this subtopic

⚠️ Where students lose marks