Home / CIE AS & A Level / CIE AS & A Level Math 9709 / CIE AS/A Level Maths-3.2 Logarithmic and exponential functions- Study Notes

CIE AS/A Level Maths-3.2 Logarithmic and exponential functions- Study Notes- New Syllabus - 2026-2027

CIE AS/A Level Maths-3.2 Logarithmic and exponential functions- Study Notes- New Syllabus

Ace AS/A Level Maths Exam with CIE AS/A Level Maths-3.2 Logarithmic and exponential functions- Study Notes

Key Concepts:

  • Relationship between Logarithms and Indices
  • The Exponential Function \(e^x\) and Natural Logarithm \(\ln x\)
  • Solving Exponential Equations and Inequalities using Logarithms
  • Using Logarithms to Linearise Relationships

AS & A Level Maths Study Notes– All Topics

Relationship between Logarithms and Indices

Relationship between Logarithms and Indices

  • A logarithm is the inverse of an exponential (index) function.
  • By definition:

    If \(a^x = N\), then \(\log_a N = x\), where \(a > 0\), \(a \neq 1\), and \(N > 0\).

  • This means logarithms “ask the question”: “To what power must I raise the base \(a\) to get \(N\)?”

Key Laws of Logarithms

  • Product Law:

    \(\log_a (MN) = \log_a M + \log_a N\)

  • Quotient Law:

    \(\log_a \left(\dfrac{M}{N}\right) = \log_a M – \log_a N\)

  • Power Law:

    \(\log_a (M^k) = k \log_a M\)

Examples:

  • \(\log_2 8 = 3\), since \(2^3 = 8\).
  • \(\log_5 (25 \times 125) = \log_5 25 + \log_5 125 = 2 + 3 = 5\).
  • \(\log_{10} \left(\dfrac{1000}{10}\right) = \log_{10} 1000 – \log_{10} 10 = 3 – 1 = 2\).
  • \(\log_3 (81^{1/2}) = \dfrac{1}{2} \log_3 81 = \dfrac{1}{2} \times 4 = 2\).

Example:

Solve for \(x\):

\(\log_2 (x+3) + \log_2 (x-1) = 3\)

▶️ Answer/Explanation

Step 1: Use the product law of logarithms.

\(\log_2 \big((x+3)(x-1)\big) = 3\)

Step 2: Rewrite the logarithm in exponential form.

\((x+3)(x-1) = 2^3\)

\((x+3)(x-1) = 8\)

Step 3: Expand and simplify.

\(x^2 + 2x – 3 = 8\)

\(x^2 + 2x – 11 = 0\)

Step 4: Solve the quadratic.

\(x = \dfrac{-2 \pm \sqrt{(2)^2 – 4(1)(-11)}}{2}\)

\(x = \dfrac{-2 \pm \sqrt{48}}{2}\)

\(x = \dfrac{-2 \pm 4\sqrt{3}}{2}\)

\(x = -1 \pm 2\sqrt{3}\)

Step 5: Check for validity (since arguments of logs must be \(> 0\)).

For \(x = -1 – 2\sqrt{3}\), the terms \((x+3)\) and \((x-1)\) are negative → not valid.

For \(x = -1 + 2\sqrt{3}\), both \((x+3) > 0\) and \((x-1) > 0\).

Final Answer:

\(\boxed{x = -1 + 2\sqrt{3}}\)

The Exponential Function \(e^x\) and Natural Logarithm \(\ln x\)

The Exponential Function \(e^x\) and Natural Logarithm \(\ln x\)

Definition:

    • \(e^x\) is the exponential function with base \(e \approx 2.718\).
    • \(\ln x\) is the natural logarithm, the inverse of \(e^x\):

\(\ln(e^x) = x\), for all real \(x\)

\(e^{\ln x} = x\), for \(x > 0\)

Properties of \(e^x\):

    • Domain: \(x \in \mathbb{R}\), Range: \((0, \infty)\)
    • Always positive: \(e^x > 0\)
    • \(e^0 = 1\)
    • Derivative: \(\dfrac{d}{dx} e^x = e^x\)

Exponential laws apply:

\(e^{a+b} = e^a \cdot e^b\)

\((e^a)^b = e^{ab}\)

Properties of \(\ln x\):

    • Domain: \(x>0\), Range: \(\mathbb{R}\)
    • \(\ln 1 = 0\)
    • Derivative: \(\dfrac{d}{dx} \ln x = \dfrac{1}{x}\)

Logarithm laws apply:

\(\ln(ab) = \ln a + \ln b\)

\(\ln(a/b) = \ln a – \ln b\)

\(\ln(a^n) = n \ln a\)

Relationship as Inverse Functions:

  • The graphs of \(y = e^x\) and \(y = \ln x\) are reflections across the line \(y=x\).
  • \(e^x\) is strictly increasing and convex; \(\ln x\) is strictly increasing and concave.
  • \(e^x\) passes through \((0,1)\); \(\ln x\) passes through \((1,0)\).

Example:

Solve for \(x\): \(e^{2x} = 7\)

▶️ Answer / Explanation

Step 1: Take natural logarithms on both sides:

\(\ln(e^{2x}) = \ln 7\)

Step 2: Use the property \(\ln(e^y) = y\):

\(2x = \ln 7\)

Step 3: Solve for \(x\):

\(x = \dfrac{\ln 7}{2}\)

Final Answer:

\(\boxed{x = \dfrac{\ln 7}{2}}\)

Example:

Solve for \(x\): \(\ln(3x-2) = 4\)

▶️ Answer / Explanation

Step 1: Rewrite in exponential form:

\(3x-2 = e^4\)

Step 2: Solve for \(x\):

\(3x = e^4 + 2 \quad \Rightarrow \quad x = \dfrac{e^4 + 2}{3}\)

Final Answer:

\(\boxed{x = \dfrac{e^4 + 2}{3}}\)

Solving Exponential Equations and Inequalities using Logarithms

Solving Exponential Equations and Inequalities using Logarithms

Key Idea:

  • If the unknown appears in the exponent, take logarithms (usually natural logarithm \(\ln\)) on both sides to bring the exponent down.
  • Use properties of logarithms: \(\ln(a^b) = b \ln a\).
  • Check the domain when solving inequalities, because logarithms are only defined for positive arguments.

Example:

Solve for \(x\): \(2^x = 5\)

▶️ Answer / Explanation

Step 1: Take natural logarithms on both sides:

\(\ln(2^x) = \ln 5\)

Step 2: Bring the exponent down using \(\ln(a^b) = b \ln a\):

\(x \ln 2 = \ln 5\)

Step 3: Solve for \(x\):

\(x = \dfrac{\ln 5}{\ln 2}\)

Final Answer:

\(\boxed{x = \dfrac{\ln 5}{\ln 2}}\)

Example:

Solve: \(3 \times 2^{3x-1} < 5\)

▶️ Answer / Explanation

Step 1: Isolate the exponential term:

\(2^{3x-1} < \dfrac{5}{3}\)

Step 2: Take natural logarithms:

\(\ln(2^{3x-1}) < \ln\left(\dfrac{5}{3}\right)\)

Step 3: Bring exponent down:

\((3x-1) \ln 2 < \ln\left(\dfrac{5}{3}\right)\)

Step 4: Solve for \(x\):

\(3x – 1 < \dfrac{\ln(5/3)}{\ln 2}\)

\(3x < 1 + \dfrac{\ln(5/3)}{\ln 2}\)

\(x < \dfrac{1}{3} + \dfrac{\ln(5/3)}{3 \ln 2}\)

Final Answer:

\(\boxed{x < \dfrac{1}{3} + \dfrac{\ln(5/3)}{3 \ln 2}}\)

Example:

Solve: \(3^{x+1} = 4^{2x-1}\)

▶️ Answer / Explanation

Step 1: Take natural logarithms on both sides:

\(\ln(3^{x+1}) = \ln(4^{2x-1})\)

Step 2: Bring exponents down:

\((x+1)\ln 3 = (2x-1)\ln 4\)

Step 3: Expand both sides:

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

Step 4: Solve for \(x\):

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

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

\(x = \dfrac{-(\ln 3 + \ln 4)}{\ln 3 – 2\ln 4} = \dfrac{\ln 12}{2\ln 4 – \ln 3}\)

Final Answer:

\(\boxed{x = \dfrac{\ln 12}{2\ln 4 – \ln 3}}\)

Using Logarithms to Linearise Relationships

Using Logarithms to Linearise Relationships

  • If a relationship involves unknown constants in an exponential or power form, taking logarithms can transform it into a straight-line (linear) equation.
  • This allows constants to be determined using the slope (gradient) and intercept from a graph of the transformed variables.

Common Transformations:

  • Power Law: \(y = k x^n \Rightarrow \ln y = \ln k + n \ln x\)

    This is linear in \(\ln x\) and \(\ln y\) with slope \(n\) and intercept \(\ln k\).

  • Exponential Law: \(y = k a^x \Rightarrow \ln y = \ln k + x \ln a\)

    This is linear in \(x\) and \(\ln y\) with slope \(\ln a\) and intercept \(\ln k\).

Example:

The relationship \(y = k x^n\) is suspected. The following data is collected:

x124816
y36122448
▶️ Answer / Explanation

Step 1: Take logarithms of both sides: \(\ln y = \ln k + n \ln x\)

Step 2: Construct table of \(\ln x\) and \(\ln y\):

x124816
y36122448
\(\ln x\)00.6931.3862.0792.773
\(\ln y\)1.0991.7922.4853.1783.871

Step 3: Plot \(\ln y\) versus \(\ln x\). The gradient = n, intercept = ln k.

Step 4: Calculate slope (n) using first and last points:

\(n = \dfrac{3.871 – 1.099}{2.773 – 0} \approx \dfrac{2.772}{2.773} \approx 1\)

Step 5: Find intercept ln k:

\(\ln k = \ln y – n \ln x = 1.099 – 1\cdot 0 = 1.099 \Rightarrow k = e^{1.099} \approx 3\)

Final Result:

\(\boxed{y \approx 3 x^1 = 3x}\)

Example:

The relationship \(y = k a^x\) is suspected. Data:

x0123
y5102040
▶️ Answer / Explanation

Step 1: Take natural logarithms: \(\ln y = \ln k + x \ln a\)

Step 2: Construct table of x vs ln y:

x0123
\(\ln y\)1.6092.3032.9963.689

Step 3: Gradient = ln a, intercept = ln k.

Gradient: \(\ln a = \dfrac{3.689 – 1.609}{3 – 0} = \dfrac{2.080}{3} \approx 0.693 \Rightarrow a = e^{0.693} \approx 2\)

Intercept: \(\ln k = 1.609 \Rightarrow k = e^{1.609} \approx 5\)

Final Result:

\(\boxed{y \approx 5 \cdot 2^x}\)

Scroll to Top