CIE AS/A Level Maths-2.2 Logarithmic and exponential functions- Study Notes- New Syllabus - 2026-2027
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Key Concepts:
- Logarithms and Indices
- The Exponential Function \(e^x\) and the Natural Logarithm \(\ln x\)
- Solving Exponential Equations and Inequalities with Logarithms
- Using Logarithms to Transform Relationships into Linear Form
Logarithms and Indices
Logarithms and Indices
Logarithms are the inverse operation of indices (exponents). They provide a way of expressing exponential equations in a different form.
Relationship between logarithms and indices
The fundamental definition is:
\(a^x = N \iff \log_a N = x\), where \(a > 0\), \(a \neq 1\), and \(N > 0\).
- If \(2^3 = 8\), then \(\log_2 8 = 3\).
- If \(10^4 = 10000\), then \(\log_{10} 10000 = 4\).
This shows that a logarithm answers the question: “To what power must the base be raised to obtain the given number?”
Basic Properties of Logarithms
| Property | Reason |
|---|---|
| \(\log_a 1 = 0\) | Because \(a^0 = 1\) |
| \(\log_a a = 1\) | Because \(a^1 = a\) |
| \(\log_a (a^k) = k\) | |
| If \(\log_a M = \log_a N\), then \(M = N\) | Logarithmic function is one-to-one |
Laws of Logarithms
| Law | Explanation |
|---|---|
| Product Law: \(\log_a (MN) = \log_a M + \log_a N\) | Follows from \(a^p \cdot a^q = a^{p+q}\) |
| Quotient Law: \(\log_a \left(\dfrac{M}{N}\right) = \log_a M – \log_a N\) | Follows from \(\dfrac{a^p}{a^q} = a^{p-q}\) |
| Power Law: \(\log_a (M^k) = k \log_a M\) | Follows from \((a^p)^k = a^{pk}\) |
Note: These are valid only for positive \(M, N\), and \(a > 0, a \neq 1\).
Example:
Express the equation \(5^3 = 125\) in logarithmic form.
▶️ Answer/Explanation
By definition: \(a^x = N \iff \log_a N = x\).
Here, \(a=5\), \(x=3\), \(N=125\).
So: \(\log_5 125 = 3\).
Example:
Simplify \(\log_2 8 + \log_2 4\).
▶️ Answer/Explanation
Apply product law: \(\log_a M + \log_a N = \log_a (MN)\).
\(\log_2 8 + \log_2 4 = \log_2 (8 \times 4) = \log_2 32\).
Since \(2^5 = 32\), \(\log_2 32 = 5\).
Final Answer: 5.
Example :
Evaluate \(\log_{10} 1000 – \log_{10} 10\).
▶️ Answer/Explanation
Apply quotient law: \(\log_a M – \log_a N = \log_a \dfrac{M}{N}\).
\(\log_{10} 1000 – \log_{10} 10 = \log_{10} \left(\dfrac{1000}{10}\right)\).
= \(\log_{10} 100\).
Since \(10^2 = 100\), this equals 2.
Final Answer: 2.
Example:
Simplify \(\log_3 81\).
▶️ Answer/Explanation
Write 81 as \(3^4\).
\(\log_3 (3^4) = 4 \log_3 3\).
\(\log_3 3 = 1\).
So the value is 4.
Final Answer: 4.
The Exponential Function \(e^x\) and the Natural Logarithm \(\ln x\)
The Exponential Function \(e^x\) and the Natural Logarithm \(\ln x\)
The number \(e\)
\(e \approx 2.71828…\) is an irrational number that arises naturally in mathematics, especially in growth and decay problems, calculus, and compound interest.
The exponential function \(e^x\)
Definition: \(e^x\) means \(e\) raised to the power \(x\).
- It is defined for all real numbers \(x\).
Key properties:
- \(e^0 = 1\).
- \(e^x > 0\) for all real \(x\).
- \(e^x\) is strictly increasing.
- \(\dfrac{d}{dx} e^x = e^x\) (it is its own derivative).
The natural logarithm \(\ln x\)
Definition: \(\ln x = \log_e x\), i.e. the logarithm with base \(e\).
- It is only defined for \(x > 0\).
Key properties:
- \(\ln 1 = 0\), since \(e^0 = 1\).
- \(\ln e = 1\), since \(e^1 = e\).
- \(\ln (ab) = \ln a + \ln b\).
- \(\ln \left(\dfrac{a}{b}\right) = \ln a – \ln b\).
- \(\ln (a^k) = k \ln a\).
- \(\dfrac{d}{dx} (\ln x) = \dfrac{1}{x}, \; x > 0.\)
Inverse relationship
\(y = e^x\) and \(y = \ln x\) are inverse functions.
This means:
- \(e^{\ln x} = x, \; x > 0\).
- \(\ln(e^x) = x, \; x \in \mathbb{R}\).
Their graphs are reflections of each other in the line \(y = x\).
Graphs
The graph of \(y = e^x\):
- Passes through \((0,1)\).
- Increases rapidly as \(x \to \infty\).
- Tends to 0 as \(x \to -\infty\), but never touches the x-axis (asymptote).
The graph of \(y = \ln x\):
- Passes through \((1,0)\).
- Increases slowly as \(x \to \infty\).
- Tends to \(-\infty\) as \(x \to 0^+\) (vertical asymptote at \(x = 0\)).
Example:
Evaluate \(\ln(e^5)\).
▶️ Answer/Explanation
Since \(\ln(e^x) = x\),
\(\ln(e^5) = 5\).
Final Answer: 5.
Example:
Solve for \(x\): \(e^x = 7\).
▶️ Answer/Explanation
Take natural logs of both sides:
\(\ln(e^x) = \ln 7\).
\(x = \ln 7\).
Final Answer: \(x = \ln 7\).
Example:
Differentiate \(y = \ln(3x^2)\).
▶️ Answer/Explanation
Apply chain rule and logarithm laws:
\(y = \ln(3) + \ln(x^2) = \ln 3 + 2 \ln x\).
\(\dfrac{dy}{dx} = 0 + \dfrac{2}{x} = \dfrac{2}{x}\).
Final Answer: \(\dfrac{dy}{dx} = \dfrac{2}{x}\).
Solving Exponential Equations and Inequalities with Logarithms
Solving Exponential Equations and Inequalities with Logarithms
General strategy
When the unknown is in the exponent (e.g. \(a^{f(x)} = b\)), we use logarithms to “bring down” the power:
- Take logarithms of both sides (usually natural log, \(\ln\), but any base works).
- Use the law \(\ln(a^k) = k \ln a\).
- Solve for the unknown.
Example forms
- Equations: \(2^x = 7\), \(3^{2x+1} = 50\), \(e^{x^2} = 10\).
- Inequalities: \(5^x > 20\), \(e^{2x} < 7\).
Important rules for inequalities
- If the base \(a > 1\), the exponential function is increasing:
- \(a^x > a^y \iff x > y\).
- If \(0 < a < 1\), the exponential function is decreasing:
- \(a^x > a^y \iff x < y\).
- For the natural exponential \(e^x\), since \(e > 1\), the function is always increasing.
Step-by-step method
- Isolate the exponential expression.
- Take natural logarithm (\(\ln\)) or logarithm of both sides.
- Bring down the exponent using \(\ln(a^k) = k \ln a\).
- Solve the resulting linear/quadratic equation or inequality.
Example:
Solve \(2^x = 7\).
▶️ Answer/Explanation
Step 1: Take natural logarithms of both sides.
\(\ln(2^x) = \ln(7)\).
Step 2: Bring down the exponent.
\(x \ln 2 = \ln 7\).
Step 3: Solve for \(x\).
\(x = \dfrac{\ln 7}{\ln 2}\).
Final Answer: \(x = \dfrac{\ln 7}{\ln 2}\).
Example:
Solve \(3^{2x+1} = 50\).
▶️ Answer/Explanation
Step 1: Take logs of both sides.
\(\ln(3^{2x+1}) = \ln(50)\).
Step 2: Bring down the exponent.
\((2x+1)\ln 3 = \ln 50\).
Step 3: Solve for \(x\).
\(2x+1 = \dfrac{\ln 50}{\ln 3}\).
\(2x = \dfrac{\ln 50}{\ln 3} – 1\).
\(x = \dfrac{1}{2}\left(\dfrac{\ln 50}{\ln 3} – 1\right)\).
Final Answer: \(x = \dfrac{\ln 50}{2\ln 3} – \dfrac{1}{2}\).
Example:
Solve \(5^x > 20\).
▶️ Answer/Explanation
Step 1: Take logarithms.
\(\ln(5^x) > \ln 20\).
Step 2: Bring down exponent.
\(x \ln 5 > \ln 20\).
Step 3: Divide through by \(\ln 5\) (positive, so inequality direction stays the same).
\(x > \dfrac{\ln 20}{\ln 5}\).
Final Answer: \(x > \dfrac{\ln 20}{\ln 5}\).
Example:
Solve \(\left(\dfrac{1}{3}\right)^x \leq \dfrac{1}{9}\).
▶️ Answer/Explanation
Step 1: Rewrite RHS as power of \(1/3\).
\(\dfrac{1}{9} = \left(\dfrac{1}{3}\right)^2\).
Step 2: Compare exponents (since base \(1/3 < 1\), function is decreasing).
\(x \geq 2\).
Final Answer: \(x \geq 2\).
Using Logarithms to Transform Relationships into Linear Form
Using Logarithms to Transform Relationships into Linear Form
The idea
Some equations are non-linear in their variables, but can be transformed into a linear form by taking logarithms. Once linearized, we can identify constants by comparing the form with the equation of a straight line:
\(Y = mX + c\), where:
- \(m\) = gradient
- \(c\) = intercept
This allows unknown constants to be determined experimentally by plotting graphs of suitable logarithmic quantities.
Two common models
(a) Power Law: \(y = kx^n\)
Take natural logarithm:
\(\ln y = \ln k + n \ln x\)
- Let \(Y = \ln y\), \(X = \ln x\).
- Equation becomes \(Y = nX + \ln k\).
- Gradient = \(n\), Intercept = \(\ln k\).
(b) Exponential Law: \(y = k a^x\)
Take natural logarithm:
\(\ln y = \ln k + x \ln a\)
- Let \(Y = \ln y\), \(X = x\).
- Equation becomes \(Y = (\ln a)X + \ln k\).
- Gradient = \(\ln a\), Intercept = \(\ln k\).
General method
- Take logarithms of both sides of the model.
- Rearrange into the form \(Y = mX + c\).
- Plot the appropriate graph (e.g. \(\ln y\) vs \(\ln x\), or \(\ln y\) vs \(x\)).
- Find gradient and intercept.
- Use them to calculate the unknown constants.
Example:
The relationship between two variables is given by \(y = kx^n\). Data is collected, and \(\ln y\) is plotted against \(\ln x\). The line of best fit has equation \(\ln y = 2.5 \ln x + 1.2\). Find \(k\) and \(n\).
▶️ Solution
Comparing with \(\ln y = n \ln x + \ln k\):
- Gradient = \(n = 2.5\).
- Intercept = \(\ln k = 1.2\).
\(k = e^{1.2} \approx 3.32\).
Final Answer: \(y \approx 3.32 x^{2.5}\).
Example:
Suppose \(y = k a^x\). A straight-line fit of \(\ln y\) against \(x\) gives line equation \(\ln y = 0.7x + 0.5\). Find \(k\) and \(a\).
▶️ Solution
Comparing with \(\ln y = (\ln a)x + \ln k\):
- Intercept = \(\ln k = 0.5 \implies k = e^{0.5} \approx 1.65\).
- Gradient = \(\ln a = 0.7 \implies a = e^{0.7} \approx 2.01\).
Final Answer: \(y \approx 1.65 \cdot (2.01)^x\).
Example:
A scientist models bacterial growth with \(y = k e^{rx}\), where \(y\) is population size after time \(x\). A graph of \(\ln y\) vs \(x\) gives a straight line with gradient \(0.4\) and intercept \(2.0\). Find \(k\) and \(r\).
▶️ Solution
Here \(y = k e^{rx}\) has log form:
\(\ln y = rx + \ln k\).
- Gradient = \(r = 0.4\).
- Intercept = \(\ln k = 2.0 \implies k = e^2 = 7.39\).
Final Answer: \(y \approx 7.39 e^{0.4x}\).