1.5, 1.7

Laws of Exponents & Rational Indices

  • Core Index Laws (for $a, b > 0$): $$a^x a^y = a^{x+y}, \qquad \frac{a^x}{a^y} = a^{x-y}, \qquad (a^x)^y = a^{xy}$$ $$(ab)^x = a^x b^x, \qquad \left(\frac{a}{b}\right)^x = \frac{a^x}{b^x}, \qquad a^{-x} = \frac{1}{a^x}, \qquad a^0 = 1$$
  • Rational Exponents and Radical Form: $$a^{1/n} = \sqrt[n]{a}, \qquad a^{m/n} = (\sqrt[n]{a})^m = \sqrt[n]{a^m} \quad (a > 0, n \in \mathbb{Z}^+)$$ Strategy: Always evaluate the root before the power to keep numbers manageable: $27^{2/3} = (\sqrt[3]{27})^2 = 3^2 = 9$.
  • Prime Base Factorization Technique: When simplifying complex exponent products or solving exponential equations, rewrite composite bases as powers of primes: $$4 = 2^2, \quad 8 = 2^3, \quad 16 = 2^4, \quad 27 = 3^3, \quad 36 = (2 \times 3)^2$$

Example 1

Simplifying Expressions with Rational Exponents

Simplify the following without using a calculator:
(a) $\frac{8^{4/3} \times 16^{-1/2}}{32^{2/5}}$
(b) $\left( \frac{27 x^6}{y^9} \right)^{-2/3}$

  1. Part (a): Express all numbers in base $2$: $$8 = 2^3 \implies 8^{4/3} = (2^3)^{4/3} = 2^4 = 16$$ $$16 = 2^4 \implies 16^{-1/2} = (2^4)^{-1/2} = 2^{-2} = \frac{1}{4}$$ $$32 = 2^5 \implies 32^{2/5} = (2^5)^{2/5} = 2^2 = 4$$ Substitute into the fraction: $$\frac{16 \times \frac{1}{4}}{4} = \frac{4}{4} = 1$$
  2. Part (b): Distribute the exponent $-2/3$ to every factor: $$\left(\frac{27 x^6}{y^9}\right)^{-2/3} = \frac{27^{-2/3} (x^6)^{-2/3}}{(y^9)^{-2/3}}$$ $$27^{-2/3} = \frac{1}{(27^{1/3})^2} = \frac{1}{3^2} = \frac{1}{9}$$ $$(x^6)^{-2/3} = x^{6 \times (-2/3)} = x^{-4} = \frac{1}{x^4}$$ $$(y^9)^{-2/3} = y^{9 \times (-2/3)} = y^{-6} = \frac{1}{y^6}$$ Combine terms: $$\frac{\frac{1}{9 x^4}}{\frac{1}{y^6}} = \frac{y^6}{9 x^4}$$

Answer: (a) $1$; (b) $\frac{y^6}{9x^4}$

1.5

Definition of a Logarithm & Domain Boundaries

  • Fundamental Equivalence: A logarithm is the power to which a base must be raised to produce a given number: $$y = a^x \iff x = \log_a y \quad (a > 0, a \neq 1, y > 0)$$
  • The Strict Domain Restriction: The argument of any logarithm must be strictly positive: $$\log_a(f(x)) \text{ is defined } \iff f(x) > 0$$ $\log_a(0)$ and $\log_a(\text{negative})$ are undefined in real numbers! Any algebraic solution that results in $f(x) \le 0$ is an extraneous root and must be rejected.
  • Fundamental Identity Laws: $$\log_a 1 = 0 \quad (\text{since } a^0 = 1), \qquad \log_a a = 1 \quad (\text{since } a^1 = a)$$ $$a^{\log_a x} = x, \qquad \log_a(a^x) = x$$
  • Natural Logarithm ($\ln$): Logarithm with base $e \approx 2.71828$: $$\ln x = \log_e x, \qquad \ln 1 = 0, \qquad \ln e = 1, \qquad e^{\ln x} = x, \qquad \ln(e^x) = x$$

Example 2

Evaluating Logarithms & Converting Forms

Evaluate without a calculator:
(a) $\log_3 81$
(b) $\log_5\left(\frac{1}{125}\right)$
(c) $\log_4 8$

  1. Part (a): $\log_3 81$
    Write $81$ as $3^4 \implies \log_3(3^4) = 4$.
  2. Part (b): $\log_5\left(\frac{1}{125}\right)$
    Write $\frac{1}{125}$ as $5^{-3} \implies \log_5(5^{-3}) = -3$.
  3. Part (c): $\log_4 8$
    Let $x = \log_4 8 \iff 4^x = 8$.
    Convert both sides to base $2$: $(2^2)^x = 2^3 \implies 2^{2x} = 2^3 \implies 2x = 3 \implies x = \frac{3}{2}$.

Answer: (a) $4$; (b) $-3$; (c) $\frac{3}{2}$

1.7

The Three Laws of Logarithms & Change of Base

  • Product Law: $$\log_a(xy) = \log_a x + \log_a y \quad (\text{Log of product = Sum of logs})$$
  • Quotient Law: $$\log_a\left(\frac{x}{y}\right) = \log_a x - \log_a y \quad (\text{Log of quotient = Difference of logs})$$
  • Power Law: $$\log_a(x^k) = k \log_a x \quad (\text{Power drops down as a multiplier})$$
  • Change of Base Rule (Formula Booklet): $$\log_b a = \frac{\log_c a}{\log_c b} = \frac{\ln a}{\ln b}$$ Allows evaluation of any logarithm on a calculator or Paper 1 using natural logs.
  • Critical Negatives & Distortions: $$\log_a(x + y) \neq \log_a x + \log_a y \quad (\text{False!})$$ $$\frac{\log_a x}{\log_a y} \neq \log_a(x - y) \quad (\text{False!})$$ $$(\log_a x)^2 \neq 2\log_a x \quad (\text{False!})$$

Example 3

Condensing & Evaluating Logarithmic Expressions

Write as a single logarithm, and hence evaluate without a calculator:
$$2\log_{10} 6 - \log_{10} 9 + \frac{1}{2}\log_{10} 625$$

  1. Apply the power law to move coefficients inside: $$2\log_{10} 6 = \log_{10}(6^2) = \log_{10} 36$$ $$\frac{1}{2}\log_{10} 625 = \log_{10}(625^{1/2}) = \log_{10} 25$$
  2. Substitute back into the expression: $$\log_{10} 36 - \log_{10} 9 + \log_{10} 25$$
  3. Apply the product and quotient laws: $$= \log_{10}\left( \frac{36 \times 25}{9} \right)$$
  4. Simplify the argument: $$\frac{36}{9} = 4 \implies 4 \times 25 = 100$$ $$= \log_{10}(100) = \log_{10}(10^2) = 2$$

Answer: $2$

1.5, 1.7

Solving Exponential & Logarithmic Equations

  • Type 1: Different Base Exponentials ($a^{f(x)} = b^{g(x)}$): Take $\ln$ of both sides, drop exponents, and solve for $x$: $$(f(x))\ln a = (g(x))\ln b$$
  • Type 2: Quadratic-in-Form Exponentials: Expressions involving $a^{2x} + p a^x + q = 0$. Substitute $u = a^x$ to produce $u^2 + p u + q = 0$. Solve for $u$, then reject any $u \le 0$ since $a^x > 0$ for all real $x$.
  • Type 3: Logarithmic Equations with Extraneous Root Checking: Combine all log terms using product/quotient laws into $\log_a(A) = \log_a(B) \implies A = B$. Mandatory Step: Test all roots in the original expressions to ensure every log argument is strictly positive!
Inverse Relationship of Exponential and Logarithm Graph showing y = e^x and y = ln(x) reflecting symmetrically across the line of symmetry y = x. x y y = x y = eˣ (0, 1) y = ln x (1, 0) Horizontal Asymptote: y = 0 Vertical Asymptote: x = 0
Figure 1.5: The exponential $y = e^x$ and logarithmic $y = \ln x$ functions are mutual inverses, reflecting across the line of symmetry $y = x$.

Example 4

Solving a Logarithmic Equation with Extraneous Root Checking

Solve the equation $\log_2(x + 5) + \log_2(x - 2) = 3$ for $x \in \mathbb{R}$.

  1. Identify initial domain restrictions: $$x + 5 > 0 \implies x > -5$$ $$x - 2 > 0 \implies x > 2$$ Combined valid domain: $x > 2$.
  2. Combine logs using the product law: $$\log_2\left( (x + 5)(x - 2) \right) = 3$$
  3. Convert to exponential form: $$(x + 5)(x - 2) = 2^3 = 8$$
  4. Expand and solve the quadratic: $$x^2 + 3x - 10 = 8 \implies x^2 + 3x - 18 = 0$$ $$(x + 6)(x - 3) = 0 \implies x = -6 \quad \text{or} \quad x = 3$$
  5. Test roots against domain ($x > 2$):
    • For $x = -6$: $\log_2(-6 - 2) = \log_2(-8)$ is undefined. Reject $x = -6$.
    • For $x = 3$: $\log_2(8) + \log_2(1) = 3 + 0 = 3$. Valid!

Answer: $x = 3$

Example 5

Quadratic in Disguise with Exponential Substitution

Solve the equation $e^{2x} - 7e^x + 10 = 0$ for $x \in \mathbb{R}$, expressing your solutions in exact form. [5 marks]

  1. Recognize the quadratic structure: Notice that $e^{2x} = (e^x)^2$. Let $u = e^x$ where $u > 0$ strictly [M1]: $$u^2 - 7u + 10 = 0 \quad [A1]$$
  2. Factorise and solve for $u$: $$(u - 2)(u - 5) = 0 \implies u = 2 \quad \text{or} \quad u = 5 \quad [M1]$$
  3. Back-substitute $u = e^x$ to find $x$: $$e^x = 2 \implies x = \ln 2 \quad [A1]$$ $$e^x = 5 \implies x = \ln 5 \quad [A1]$$

Answers: $x = \ln 2$ or $x = \ln 5$

Examiner Commentary: Both values of $u$ are positive, so both yield valid real solutions. If one value had been negative (e.g. $u = -3$), candidates must explicitly reject it since $e^x > 0$ for all real $x$.

Tier 1: Core Fluency (Questions 1–4)
Exercise 1 Paper 1 Core Exponent Evaluation

Evaluate the following expressions without using a calculator:

(a) $64^{2/3}$

(b) $\left(\frac{16}{81}\right)^{-3/4}$

View Complete Solution
  1. Part (a): $$64^{2/3} = (\sqrt[3]{64})^2 = 4^2 = 16$$
  2. Part (b): $$\left(\frac{16}{81}\right)^{-3/4} = \left(\frac{81}{16}\right)^{3/4} = \left(\sqrt[4]{\frac{81}{16}}\right)^3 = \left(\frac{3}{2}\right)^3 = \frac{27}{8}$$

Final Answer: (a) $16$; (b) $\frac{27}{8}$

Exercise 2 Paper 1 Basic Log Laws

Evaluate without a calculator: $\log_2 32 - \log_3\left(\frac{1}{27}\right) + \ln(e^5)$.

View Complete Solution
  1. Evaluate each term separately: $$\log_2 32 = \log_2(2^5) = 5$$ $$\log_3\left(\frac{1}{27}\right) = \log_3(3^{-3}) = -3$$ $$\ln(e^5) = 5$$
  2. Combine terms: $$5 - (-3) + 5 = 5 + 3 + 5 = 13$$

Final Answer: $13$

Exercise 3 Paper 1 Solving Base-Equated Exponential

Solve for $x$: $4^{2x - 1} = 32$.

View Complete Solution
  1. Convert both sides to base $2$: $$4 = 2^2 \implies (2^2)^{2x - 1} = 2^{2(2x - 1)} = 2^{4x - 2}$$ $$32 = 2^5$$
  2. Equate the exponents: $$4x - 2 = 5 \implies 4x = 7 \implies x = \frac{7}{4}$$

Final Answer: $x = \frac{7}{4}$

Exercise 4 Paper 1 Condensing Log Expression

Write the expression $3\ln x - 2\ln y + \frac{1}{2}\ln z$ as a single logarithm.

View Complete Solution
  1. Apply power law to all coefficients: $$3\ln x = \ln(x^3), \quad 2\ln y = \ln(y^2), \quad \frac{1}{2}\ln z = \ln(z^{1/2}) = \ln(\sqrt{z})$$
  2. Combine using product and quotient laws: $$\ln(x^3) - \ln(y^2) + \ln(\sqrt{z}) = \ln\left( \frac{x^3 \sqrt{z}}{y^2} \right)$$

Final Answer: $\ln\left( \frac{x^3 \sqrt{z}}{y^2} \right)$

Tier 2: Exam Modeling (Questions 5–7)
Exercise 5 Paper 1 Quadratic-in-Form Exponential

Solve the equation $9^x - 10(3^x) + 9 = 0$ for $x \in \mathbb{R}$.

View Complete Solution
  1. Rewrite $9^x$ as $(3^2)^x = (3^x)^2$: $$(3^x)^2 - 10(3^x) + 9 = 0$$
  2. Let $u = 3^x > 0$: $$u^2 - 10u + 9 = 0$$
  3. Factor the quadratic: $$(u - 1)(u - 9) = 0 \implies u = 1 \quad \text{or} \quad u = 9$$
  4. Substitute back $u = 3^x$: $$3^x = 1 \implies x = 0$$ $$3^x = 9 = 3^2 \implies x = 2$$

Final Answer: $x = 0 \quad \text{and} \quad x = 2$

Exercise 6 Paper 1 Log Equation with Extraneous Root

Solve the equation $\log_3(x + 2) + \log_3(x - 6) = 2$.

View Complete Solution
  1. State domain: $x + 2 > 0 \implies x > -2$ and $x - 6 > 0 \implies x > 6$. Combined domain: $x > 6$.
  2. Combine into single logarithm: $$\log_3\left( (x + 2)(x - 6) \right) = 2$$
  3. Convert to exponential form: $$(x + 2)(x - 6) = 3^2 = 9$$ $$x^2 - 4x - 12 = 9 \implies x^2 - 4x - 21 = 0$$
  4. Factor: $$(x - 7)(x + 3) = 0 \implies x = 7 \quad \text{or} \quad x = -3$$
  5. Domain check ($x > 6$): $x = -3$ yields $\log_3(-1)$, which is undefined. Reject $x = -3$.
    Therefore, $x = 7$.

Final Answer: $x = 7$

Exercise 7 Paper 1 Variable Substitution with Log Laws

Given that $\log_2 3 = p$ and $\log_2 5 = q$, express $\log_2 45$ in terms of $p$ and $q$.

View Complete Solution
  1. Factor $45$ into prime factors: $$45 = 9 \times 5 = 3^2 \times 5$$
  2. Apply product and power laws: $$\log_2 45 = \log_2(3^2 \times 5) = \log_2(3^2) + \log_2 5 = 2\log_2 3 + \log_2 5$$
  3. Substitute $p$ and $q$: $$\log_2 45 = 2p + q$$

Final Answer: $2p + q$

Tier 3: Complex Applications & IB Exam Modeling (Questions 8–10)
Exercise 8 Paper 1 Different Bases Exponential

Solve the equation $3^{x+1} = 5^x$, giving your answer in the form $\frac{\ln a}{\ln b}$ where $a, b \in \mathbb{Q}$.

View Complete Solution
  1. Take natural logarithms of both sides: $$\ln(3^{x+1}) = \ln(5^x)$$ $$(x + 1)\ln 3 = x \ln 5$$
  2. Expand and collect terms in $x$: $$x \ln 3 + \ln 3 = x \ln 5$$ $$x \ln 5 - x \ln 3 = \ln 3$$ $$x(\ln 5 - \ln 3) = \ln 3$$
  3. Use quotient law on the bracket: $$\ln 5 - \ln 3 = \ln\left(\frac{5}{3}\right)$$ $$x = \frac{\ln 3}{\ln(5/3)}$$

Final Answer: $x = \frac{\ln 3}{\ln(5/3)}$

Exercise 9 Paper 1 Simultaneous System

Solve the simultaneous equations for $x$ and $y$:

$$\begin{cases} 2^x \times 4^y = 128 \\ \log_3(x + y) = 2 \end{cases}$$

View Complete Solution
  1. Simplify the first equation in base $2$: $$2^x \times (2^2)^y = 2^x \times 2^{2y} = 2^{x + 2y}$$ $$128 = 2^7 \implies x + 2y = 7 \quad \text{--- (1)}$$
  2. Simplify the second equation in exponential form: $$\log_3(x + y) = 2 \implies x + y = 3^2 = 9 \quad \text{--- (2)}$$
  3. Subtract (2) from (1): $$(x + 2y) - (x + y) = 7 - 9 \implies y = -2$$
  4. Substitute $y = -2$ into (2): $$x + (-2) = 9 \implies x = 11$$
  5. Check domain: $x + y = 11 - 2 = 9 > 0$ (valid).

Final Answer: $x = 11, y = -2$

Exercise 10 Paper 1 Multi-part Function & Inverse

Let $f(x) = \ln(2x - 5)$ for $x > 2.5$.

(a) Write down the range of $f$.

(b) Find an expression for $f^{-1}(x)$, stating its domain.

(c) Solve the equation $f(x) = \ln(x + 1) + \ln 3$.

View Complete Solution
  1. Part (a): The range of any natural log function with an open linear argument is $\mathbb{R}$ (all real numbers).
  2. Part (b): Find inverse function: Let $y = \ln(2x - 5)$. Swap variables: $$x = \ln(2y - 5) \implies e^x = 2y - 5 \implies 2y = e^x + 5 \implies y = \frac{e^x + 5}{2}$$ $$f^{-1}(x) = \frac{e^x + 5}{2}, \quad \text{Domain: } x \in \mathbb{R}$$
  3. Part (c): Solve $f(x) = \ln(x + 1) + \ln 3$: $$\ln(2x - 5) = \ln(3(x + 1))$$ $$2x - 5 = 3x + 3$$ $$2x - 3x = 3 + 5 \implies -x = 8 \implies x = -8$$ Domain check: For $x = -8$, $2(-8) - 5 = -21 \le 0$ and $-8 + 1 = -7 \le 0$. Both are undefined in $\mathbb{R}$.
    Therefore, there are no real solutions.

Final Answer: (a) $y \in \mathbb{R}$; (b) $f^{-1}(x) = \frac{e^x + 5}{2}, x \in \mathbb{R}$; (c) No real solutions

Formula Booklet SL 1.5, 1.7

Laws of Logarithms

  • Product Law: $$\log_a(xy) = \log_a x + \log_a y$$
  • Quotient Law: $$\log_a\left(\frac{x}{y}\right) = \log_a x - \log_a y$$
  • Power Law: $$\log_a(x^k) = k \log_a x$$
  • Change of Base: $$\log_b a = \frac{\log_c a}{\log_c b} = \frac{\ln a}{\ln b}$$
Must Memorize

Laws of Exponents

  • Product & Quotient: $$a^x a^y = a^{x+y}, \qquad \frac{a^x}{a^y} = a^{x-y}$$
  • Power of a Power: $$(a^x)^y = a^{xy}$$
  • Rational Exponent: $$a^{m/n} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m$$
  • Negative Exponent: $$a^{-x} = \frac{1}{a^x}$$
Core Identities

Logarithmic Identities & Domains

  • Base Equivalence: $$y = a^x \iff x = \log_a y$$
  • Natural Log Base $e$: $$\ln(e) = 1, \quad \ln(1) = 0, \quad e^{\ln x} = x$$
  • Strict Domain Restriction: $$\log_a(g(x)) \implies g(x) > 0$$

Critical IB Examination Pitfalls in Logarithms

  • Forgetting Extraneous Root Checking: When solving $\log(A) + \log(B) = C$, you create a quadratic equation that often gives two roots. You MUST test both in the original expressions; roots making any argument $\le 0$ must be rejected!
  • Distributing Logarithms Over Addition: $\log(x + y)$ is completely simplified. Writing $\log(x + y) = \log x + \log y$ is mathematically invalid.
  • Confusing Power Rule with Whole-Log Powers: $\log(x^2) = 2\log x$, BUT $(\log x)^2 = (\log x)(\log x) \neq 2\log x$.
  • Dividing Logs vs Quotient Law: $\frac{\log x}{\log y} \neq \log(x - y)$. It equals $\log_y x$ by change of base.
  • Negative Base Exponents: $(-2)^x$ is not an exponential function over real numbers. Base $a$ must always satisfy $a > 0$ and $a \neq 1$.

Section A: Diagnostic Check

4 core concept check questions. Immediate feedback and scoring.

Score: 0 / 4

Question 1

What is the exact value of $27^{-4/3}$?

Correct! $27^{-4/3} = \frac{1}{(27^{1/3})^4} = \frac{1}{3^4} = \frac{1}{81}$.

Question 2

If $\log_3(x - 2) = 4$, what is the value of $x$?

Correct! $x - 2 = 3^4 = 81 \implies x = 81 + 2 = 83$.

Question 3

Which expression is mathematically equivalent to $\log_2 10$?

Correct! By the change of base theorem, $\log_b a = \frac{\ln a}{\ln b}$, so $\log_2 10 = \frac{\ln 10}{\ln 2}$.

Question 4

For what domain of $x$ is the function $f(x) = \ln(6 - 2x)$ defined?

Correct! The argument must be strictly positive: $6 - 2x > 0 \implies 2x < 6 \implies x < 3$.

Section B: Examination Practice

Multi-step questions with markscheme breakdowns.

Paper 1 (Non-Calculator) [6 marks]

Consider the equation $4^x - 6(2^x) + 8 = 0$.

(a) Show that the equation can be written as $u^2 - 6u + 8 = 0$, where $u = 2^x$. [2 marks]
(b) Hence, find all real values of $x$. [4 marks]
View Markscheme
(a)
Rewrite $4^x = (2^2)^x = (2^x)^2$ M1
Substituting $u = 2^x$ gives $u^2 - 6u + 8 = 0$ A1

(b)
Factor quadratic: $(u - 2)(u - 4) = 0$ M1
$u = 2 \quad \text{or} \quad u = 4$ A1
Substitute back $2^x = 2 \implies \mathbf{x = 1}$ A1
$2^x = 4 = 2^2 \implies \mathbf{x = 2}$ A1
Paper 2 (Calculator Allowed) [6 marks]

A radioactive isotope decays according to the equation $N(t) = 450 e^{-0.035 t}$, where $N$ is the mass in grams remaining after $t$ days.

(a) State the initial mass of the isotope. [1 mark]
(b) Find the mass remaining after $10$ days, correct to $3$ significant figures. [2 marks]
(c) Find the half-life of the isotope (the time taken for the mass to reduce to half of its initial value), giving your answer in days to $3$ significant figures. [3 marks]
View Markscheme
(a)
At $t = 0$: $N(0) = 450 e^0 = \mathbf{450\text{ grams}}$ A1

(b)
$N(10) = 450 e^{-0.035(10)} = 450 e^{-0.35}$ M1
$= 450(0.704688) \approx \mathbf{317\text{ grams}}$ A1

(c)
Set $N(t) = 225 \implies 225 = 450 e^{-0.035 t} \implies e^{-0.035 t} = 0.5$ M1
$-0.035 t = \ln(0.5) = -\ln 2$ M1
$t = \frac{\ln 2}{0.035} \approx \frac{0.693147}{0.035} \approx \mathbf{19.8\text{ days}}$ A1

Unit Overview

All Lessons in Number & Algebra (Topic 1)

1.1 Standard Form Scientific notation, powers-of-ten operations, and the IB 3 s.f. rule. 1.2 Arithmetic Sequences & Series Common difference, general $n$th term, and partial sum formulas. 1.3, 1.8 Geometric Sequences & Infinite Sums Common ratio, finite sums, and convergence to infinity when $|r| < 1$. 1.4 Financial Applications Compound interest, annual depreciation, and TVM solver modeling. 1.5, 1.7 Exponents & Logarithms Laws of exponents, logarithm rules, base changes, and natural logs. 1.6, 1.9 Proof & Binomial Theorem Deductive proof, Pascal's triangle, and binomial expansion terms.