Logarithmic and exponential functions

1. Overview

Logarithmic and exponential functions are fundamental in Additional Mathematics. They are inverses of each other and are crucial for modelling various real-world phenomena. Mastering these functions is essential for success in both Paper 1 (non-calculator) and Paper 2 (calculator) exams. Expect to encounter questions involving simplifying expressions using log laws, solving exponential equations, and understanding the properties of their graphs. This revision note covers key definitions, laws, and techniques, with a focus on avoiding common mistakes and maximising your exam performance.

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Key Definitions

• Exponential Function: A function of the form $f(x) = a^x$, where $a > 0$. The most important base is Euler’s number $e \approx 2.718$.
• Natural Logarithm ($\ln x$): A logarithm to the base $e$. It is the inverse of the exponential function $e^x$.
• Common Logarithm ($\lg x$): A logarithm to the base 10.
• Asymptote: A line that a graph approaches infinitely closely but never touches.
• Extraneous Solution: A solution that emerges from the algebraic process but is invalid because it falls outside the domain of the original function (e.g., $\ln(-2)$ is undefined).

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Core Content

3.1 The Graphs of $e^x$ and $\ln x$

• $y = e^x$: Passes through $(0, 1)$. It has a horizontal asymptote at $y = 0$. The domain is $x \in \mathbb{R}$ and the range is $y > 0$.
• $y = \ln x$: Passes through $(1, 0)$. It has a vertical asymptote at $x = 0$. The domain is $x > 0$ and the range is $y \in \mathbb{R}$.
• Inverse Relationship: $f(x) = e^x$ and $g(x) = \ln x$ are reflections of each other in the line $y = x$.
  • $e^{\ln x} = x$
  • $\ln(e^x) = x$

Transformed Graphs:

• For $y = ke^{nx} + a$: The horizontal asymptote shifts to $y = a$.
• For $y = k \ln(ax + b)$: The vertical asymptote is found by setting $ax + b = 0$.

3.2 Laws of Logarithms

For any positive base $a$ ($a \neq 1$):

1. Product Rule: $\log_a(xy) = \log_a x + \log_a y$
2. Quotient Rule: $\log_a(\frac{x}{y}) = \log_a x - \log_a y$
3. Power Rule: $\log_a(x^n) = n \log_a x$
4. Change of Base: $\log_a b = \frac{\log_c b}{\log_c a}$ (Useful for Paper 2 calculator questions).
5. Special Cases: $\log_a a = 1$ and $\log_a 1 = 0$.

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Worked Example 1 — Simplifying Logarithmic Expressions

Express $2\ln x - \frac{1}{3}\ln y + 4$ as a single logarithm.

Step 1: Apply the power rule to the first two terms.
$2\ln x = \ln(x^2)$
$\frac{1}{3}\ln y = \ln(y^{\frac{1}{3}}) = \ln(\sqrt[3]{y})$

Step 2: Express the constant 4 as a natural logarithm.
$4 = 4\ln e = \ln(e^4)$
Reason: Since $\ln e = 1$, we can multiply by 4 and then use the power rule.

Step 3: Combine the logarithmic terms using the product and quotient rules.
$\ln(x^2) - \ln(\sqrt[3]{y}) + \ln(e^4) = \ln\left(\frac{x^2}{\sqrt[3]{y}}\right) + \ln(e^4) = \ln\left(\frac{e^4x^2}{\sqrt[3]{y}}\right)$
Reason: Quotient rule, then product rule.

Final Answer: $\boxed{\ln\left(\frac{e^4x^2}{\sqrt[3]{y}}\right)}$

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Worked Example 2 — Solving Exponential Equations

Solve the equation $3^{x+2} = 5^{2x-1}$ for $x$. Give your answer in exact form.

Step 1: Take the natural logarithm of both sides.
$\ln(3^{x+2}) = \ln(5^{2x-1})$
Reason: Applying the natural logarithm to both sides allows us to use the power rule.

Step 2: Apply the power rule to both sides.
$(x+2)\ln 3 = (2x-1)\ln 5$

Step 3: Expand the brackets.
$x\ln 3 + 2\ln 3 = 2x\ln 5 - \ln 5$
Reason: Expanding to isolate x terms.

Step 4: Rearrange to group terms with $x$ on one side.
$2\ln 3 + \ln 5 = 2x\ln 5 - x\ln 3$

Step 5: Factor out $x$.
$2\ln 3 + \ln 5 = x(2\ln 5 - \ln 3)$

Step 6: Solve for $x$.
$x = \frac{2\ln 3 + \ln 5}{2\ln 5 - \ln 3}$
Reason: Dividing to isolate x.

Step 7: Simplify using log rules (optional, but good practice for Paper 1).
$x = \frac{\ln(3^2) + \ln 5}{\ln(5^2) - \ln 3} = \frac{\ln 9 + \ln 5}{\ln 25 - \ln 3} = \frac{\ln(45)}{\ln(\frac{25}{3})}$
Reason: Applying power rule and then product/quotient rules.

Final Answer: $\boxed{x = \frac{\ln 45}{\ln \frac{25}{3}}}$

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Worked Example 3 — Solving Equations with Substitution

Solve the equation $e^{2x} - 7e^x + 12 = 0$.

Step 1: Substitute $u = e^x$. This means $u^2 = e^{2x}$.
Reason: Simplifying the equation into a quadratic form.

Step 2: Rewrite the equation in terms of $u$.
$u^2 - 7u + 12 = 0$

Step 3: Factorise the quadratic equation.
$(u - 3)(u - 4) = 0$
Reason: Factorising to find the roots.

Step 4: Solve for $u$.
$u = 3$ or $u = 4$

Step 5: Substitute back $e^x$ for $u$ and solve for $x$.
$e^x = 3 \implies x = \ln 3$
$e^x = 4 \implies x = \ln 4$

Final Answer: $\boxed{x = \ln 3, \ln 4}$

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Worked Example 4 — Solving Logarithmic Equations

Solve the equation $\log_2(x+2) + \log_2(x-1) = 2$.

Step 1: Combine the logarithms using the product rule.
$\log_2((x+2)(x-1)) = 2$

Step 2: Expand the expression inside the logarithm.
$\log_2(x^2 + x - 2) = 2$

Step 3: Convert the logarithmic equation to exponential form.
$x^2 + x - 2 = 2^2$
Reason: Definition of logarithm.

Step 4: Simplify and rearrange to form a quadratic equation.
$x^2 + x - 2 = 4$
$x^2 + x - 6 = 0$

Step 5: Factorise the quadratic equation.
$(x+3)(x-2) = 0$
Reason: Factorising to find the roots.

Step 6: Solve for $x$.
$x = -3$ or $x = 2$

Step 7: Check for extraneous solutions.
If $x = -3$, then $\log_2(-3+2) = \log_2(-1)$, which is undefined. Therefore, $x = -3$ is an extraneous solution.
If $x = 2$, then $\log_2(2+2) = \log_2(4)$ and $\log_2(2-1) = \log_2(1)$, which are both defined.

Final Answer: $\boxed{x = 2}$

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In More Depth

Additional Mathematics is a single-tier syllabus — all content above applies to all students.

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Key Equations

$\log_a x = y \iff a^y = x$ (Definition of a logarithm)

$\log_a b = \frac{\ln b}{\ln a}$ (Change of base formula, not on the 0606 formula sheet)

$x^{-n} = \frac{1}{x^n}$ (Negative exponent rule, not on the 0606 formula sheet)

$\ln e = 1$ and $\ln 1 = 0$ (Key natural log values)

$\frac{1}{\log_a b} = \log_b a$ (Reciprocal log identity)

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Common Mistakes to Avoid

• ❌ Wrong: $\ln(A + B) = \ln A + \ln B$
  ✓ Right: There is no rule to expand the log of a sum. Only $\ln(AB) = \ln A + \ln B$.
• ❌ Wrong: Retaining $x = -2$ as a solution for $\ln(x)$.
  ✓ Right: Always check if your solution is within the domain. You cannot take the log of a negative number or zero.
• ❌ Wrong: Writing $\frac{\ln A}{\ln B}$ as $\ln(A - B)$.
  ✓ Right: $\frac{\ln A}{\ln B}$ is the change of base formula; $\ln(\frac{A}{B}) = \ln A - \ln B$.
• ❌ Wrong: In substitution, forgetting to solve for the original variable $x$ after finding $u$.
  ✓ Right: Always substitute back (e.g., $e^x = u$) to find the final value of $x$.
• ❌ Wrong: Giving a decimal approximation when an exact answer (in terms of $\ln$ or a surd) is required.
  ✓ Right: Unless the question explicitly asks for a decimal answer to a certain number of significant figures, leave your answer in exact form. For example, write $\ln 5$ instead of $1.609$.
• ❌ Wrong: Forgetting to check for extraneous solutions when solving logarithmic equations.
  ✓ Right: Always substitute your solutions back into the original equation to ensure that you are not taking the logarithm of a negative number or zero.
• ❌ Wrong: Making sign errors when rearranging equations involving logarithms.
  ✓ Right: Be meticulous with your algebraic manipulations, paying close attention to signs, especially when expanding brackets or moving terms across the equals sign.
• ❌ Wrong: Incorrectly applying the power rule of logarithms, e.g., writing $(\ln x)^2$ as $2\ln x$.
  ✓ Right: Remember that $\ln(x^2) = 2\ln x$, but $(\ln x)^2$ is simply $(\ln x) \times (\ln x)$.

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Exam Tips

1. Exact Values: Unless the question asks for "3 significant figures," always leave your answer in exact form (e.g., $\ln 3$ or $\frac{1 + \sqrt{5}}{2}$).
2. Show Your Steps: In "Show that" questions or Paper 1 (non-calculator), explicitly show the application of log laws. Jumping from $2 \lg p - \lg q$ to $\lg(\frac{p^2}{q})$ without an intermediate step can lose marks.
3. Command Words:
   • "Solve": Find the value of the variable.
   • "Express as a single logarithm": Use log laws to condense the expression.
   • "Find the exact coordinates": Use $\ln$ or $e$ to find points where a graph crosses axes.
4. Substitution Strategy: If you see an equation with both $e^{2x}$ and $e^x$ (or $3^{2x}$ and $3^x$), it is almost always a quadratic in disguise. State your substitution clearly (e.g., "Let $u = e^x$").
5. Domain Restrictions: When solving $\log_2(x+3) + \log_2(x) = \dots$, if you get solutions $x=2$ and $x=-5$, you must reject $x=-5$ because it would result in $\log_2(-5)$, which is undefined.
6. Paper 1 Focus: Paper 1 requires strong algebraic skills. Practice simplifying logarithmic expressions and solving equations without a calculator. Pay close attention to exact values.
7. Paper 2 Advantage: Use your calculator effectively in Paper 2 for change of base calculations and to check your answers. However, still show your working, especially in "show that" questions.

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0606 papers.

Exam-Style Question 1 — Paper 1 (No Calculator Allowed) [7 marks]

Question:

(a) Solve the equation $2^{2x+1} = 9 \times 2^x$. [4]

(b) Find the value of $x$ for which $\log_3(x+2) + \log_3(x-2) = 2$. [3]

Worked Solution:

(a)

1. Rewrite the equation using laws of indices:
   $2^{2x} \times 2^1 = 9 \times 2^x$
   Separating the powers

2. Rearrange into a quadratic form:
   $2(2^x)^2 - 9(2^x) = 0$

3. Factorise:
   $2^x(2(2^x) - 9) = 0$
   Factoring out $2^x$

4. Solve for $2^x$:
   $2(2^x) - 9 = 0$, so $2^x = \frac{9}{2}$
   Since $2^x$ cannot be 0

5. Take logarithms base 2:
   $x = \log_2(\frac{9}{2}) = \log_2 9 - \log_2 2 = \log_2 9 - 1$
   Applying log rules

6. Final answer:
   $\boxed{x = \log_2 9 - 1}$

How to earn full marks: Show each step of algebraic manipulation clearly, especially when rearranging and factoring. Remember to use logarithm rules correctly.

(b)

1. Combine the logarithms:
   $\log_3((x+2)(x-2)) = 2$
   Using the log rule $\log a + \log b = \log ab$

2. Simplify:
   $\log_3(x^2 - 4) = 2$
   Expanding the bracket

3. Convert to exponential form:
   $x^2 - 4 = 3^2 = 9$

4. Solve for $x^2$:
   $x^2 = 13$
   Adding 4 to both sides

5. Solve for $x$:
   $x = \pm \sqrt{13}$
   Taking the square root

6. Check for validity:
   $x = -\sqrt{13}$ is not valid as it leads to the logarithm of a negative number.

7. Final answer:
   $\boxed{x = \sqrt{13}}$

How to earn full marks: Remember to check for extraneous solutions after solving logarithmic equations. Clearly state why you are rejecting any invalid solutions.

Common Pitfall: Remember to always check your solutions when dealing with logarithms. Negative values inside a logarithm are undefined, so you must discard any solutions that lead to this.

Exam-Style Question 2 — Paper 1 (No Calculator Allowed) [8 marks]

Question:

(a) Given that $\log_b x = p$ and $\log_b y = q$, express $\log_b(\frac{x^3}{y^2})$ in terms of $p$ and $q$. [2]

(b) Solve for $x$ in the equation $\log_2(x+3) - \log_2(3x-7) = 1$. [4]

(c) Find the exact value of $x$ satisfying $e^{2x} - 4e^x + 3 = 0$. [2]

Worked Solution:

(a)

1. Apply the laws of logarithms:
   $\log_b(\frac{x^3}{y^2}) = \log_b x^3 - \log_b y^2$
   Using the quotient rule for logarithms

2. Simplify further:
   $3 \log_b x - 2 \log_b y = 3p - 2q$
   Using the power rule for logarithms and substituting p and q

3. Final answer:
   $\boxed{3p - 2q}$

How to earn full marks: Apply the logarithm rules correctly and substitute the given variables accurately. Show each step clearly.

(b)

1. Combine the logarithms:
   $\log_2(\frac{x+3}{3x-7}) = 1$
   Using the quotient rule for logarithms

2. Convert to exponential form:
   $\frac{x+3}{3x-7} = 2^1 = 2$

3. Solve for $x$:
   $x+3 = 2(3x-7)$
   Multiplying both sides by (3x-7)

4. Expand and rearrange:
   $x+3 = 6x - 14$

5. Simplify:
   $5x = 17$
   Rearranging for x

6. Final answer:
   $\boxed{x = \frac{17}{5}}$

How to earn full marks: Remember to combine the logarithms into a single term before converting to exponential form. Show all algebraic steps clearly.

(c)

1. Recognize the quadratic form:
   $(e^x)^2 - 4e^x + 3 = 0$

2. Factorise the quadratic:
   $(e^x - 1)(e^x - 3) = 0$

3. Solve for $e^x$:
   $e^x = 1$ or $e^x = 3$

4. Solve for $x$:
   $x = \ln 1 = 0$ or $x = \ln 3$
   Taking natural logarithms

5. State the answer:
   $\boxed{x = 0, \ln 3}$

How to earn full marks: Recognize the quadratic form and factorise correctly. Remember that $\ln(1) = 0$.

Common Pitfall: When solving equations involving $e^x$, remember that $e^0 = 1$, so $\ln 1 = 0$. Don't forget this simple solution!

Exam-Style Question 3 — Paper 2 (Calculator Allowed) [7 marks]

Question:

(a) Solve the equation $4^{2x-1} = 9^{x+2}$, giving your answer correct to 3 significant figures. [4]

(b) The variables $x$ and $y$ are related by the equation $y = Ae^{kx}$, where $A$ and $k$ are constants. When $x=1$, $y=8$ and when $x=4$, $y=20$. Find the values of $A$ and $k$, correct to 3 significant figures. [3]

Worked Solution:

(a)

1. Take logarithms of both sides:
   $\ln(4^{2x-1}) = \ln(9^{x+2})$

2. Apply the power rule of logarithms:
   $(2x-1)\ln 4 = (x+2)\ln 9$

3. Expand the brackets:
   $2x\ln 4 - \ln 4 = x\ln 9 + 2\ln 9$

4. Rearrange to solve for $x$:
   $2x\ln 4 - x\ln 9 = 2\ln 9 + \ln 4$

5. Factorise and solve for $x$:
   $x(2\ln 4 - \ln 9) = 2\ln 9 + \ln 4$
   Factoring out $x$

6. Final value:
   $x = \frac{2\ln 9 + \ln 4}{2\ln 4 - \ln 9}$
   Solving for $x$

7. Calculate the final numerical value:
   $x = \frac{2\ln 9 + \ln 4}{2\ln 4 - \ln 9} = \frac{5.7807...}{0.57536...} \approx 10.047$

8. State the answer:
   $\boxed{x = 10.0}$

How to earn full marks: Show all steps of algebraic manipulation, especially when rearranging to isolate $x$. Give the final answer to the specified number of significant figures.

(b)

1. Substitute the given values into the equation:
   $8 = Ae^{k}$ and $20 = Ae^{4k}$

2. Divide the second equation by the first equation:
   $\frac{20}{8} = \frac{Ae^{4k}}{Ae^{k}}$

3. Simplify:
   $2.5 = e^{3k}$

4. Take natural logarithms:
   $\ln 2.5 = 3k$

5. Solve for $k$:
   $k = \frac{\ln 2.5}{3} \approx 0.305$

6. Substitute $k$ back into one of the original equations to find $A$:
   $8 = Ae^{0.30543...}$ (use the unrounded value of $k$)

7. Solve for $A$:
   $A = \frac{8}{e^{0.30543...}} \approx 5.89$

8. State the answers:
   $\boxed{A = 5.89, k = 0.305}$

How to earn full marks: Show the substitution of values and the division of equations clearly. Give both $A$ and $k$ to the specified number of significant figures.

Common Pitfall: When solving simultaneous equations with exponentials, dividing one equation by the other is a useful trick to eliminate one of the variables. Make sure you divide the equations in the correct order to avoid negative exponents.

Exam-Style Question 4 — Paper 2 (Calculator Allowed) [9 marks]

Question:

(a) Given that $y = 5 \ln(3x+2) - 6$, find the value of $x$ when $y=4$. Give your answer correct to 3 significant figures. [3]

(b) (i) Sketch the graph of $y = 3e^{0.4x} - 2$ for $-5 \leq x \leq 5$, showing the coordinates of any points where the curve crosses the axes. [4]
(ii) Use your graph to estimate the solution to the equation $3e^{0.4x} = 4$. [2]

Worked Solution:

(a)

1. Substitute $y=4$ into the equation:
   $4 = 5 \ln(3x+2) - 6$

2. Rearrange the equation:
   $10 = 5 \ln(3x+2)$

3. Divide by 5:
   $2 = \ln(3x+2)$

4. Convert to exponential form:
   $e^2 = 3x+2$

5. Solve for $x$:
   $3x = e^2 - 2$
   Rearranging

6. Final value of x:
   $x = \frac{e^2 - 2}{3} \approx 1.796$
   Solving for $x$

7. State the answer:
   $\boxed{x = 1.80}$

How to earn full marks: Show each step of rearranging the equation and converting to exponential form. Give the final answer to 3 significant figures.

(b)
(i)

1. Find the y-intercept:
   When $x=0$, $y = 3e^{0.4(0)} - 2 = 3(1) - 2 = 1$. The y-intercept is $(0, 1)$.

2. Find the x-intercept:
   When $y=0$, $3e^{0.4x} - 2 = 0$, so $3e^{0.4x} = 2$, $e^{0.4x} = \frac{2}{3}$, $0.4x = \ln \frac{2}{3}$, $x = \frac{\ln \frac{2}{3}}{0.4} \approx -1.01$. The x-intercept is approximately $(-1.01, 0)$.

3. Sketch the graph showing the y-intercept at (0,1) and the x-intercept at approximately (-1.01, 0). The graph is an increasing exponential curve.

4. Coordinates of intercepts:
   x-intercept: $(-1.01, 0)$
   y-intercept: $(0, 1)$

How to earn full marks: Calculate the x and y intercepts accurately and label them clearly on your sketch. Make sure the shape of the exponential curve is correct.

(ii)

1. Rewrite the equation:
   $3e^{0.4x} -2 = 4 -2$
   subtracting 2 from both sides

2. Read the value of $x$ from the graph where $y = 4 - 2 = 2$.

3. From the graph, $x \approx 0.7$.

4. State the answer:
   $\boxed{x = 0.7}$

How to earn full marks: Show how you are using the graph to find the solution. Draw a line on the graph at y=2 and read off the corresponding x-value.

Common Pitfall: When estimating solutions from a graph, make sure you read the axes carefully and provide a reasonable approximation. Don't just guess a number – show how you're using the graph to find your answer.