A-Level Mathematics (9709) P3 Study Notes: Logarithmic and Exponential Functions
Welcome to one of the most powerful and fundamental topics in Pure Mathematics 3! Logarithmic and exponential functions (often called 'Log and Exp') are the mathematical tools used to model growth and decay in the real world—from compound interest and radioactive decay to population growth and cooling rates.
In this chapter, we will master the rules that govern these functions, especially focusing on the natural base, \(e\), and its corresponding logarithm, \(\ln x\). Mastering this topic is essential, as these functions form the basis for much of the calculus you will study later in P3 (Differentiation and Integration).
1. The Fundamental Relationship: Indices and Logarithms
Logarithms are simply a way of expressing indices (or powers). They answer the question: "What power do I raise the base to, to get a specific number?"
Key Definition (The Log Switch)
If we have an exponential statement:
\[
\mathbf{a}^x = \mathbf{N}
\]
This can be converted directly into a logarithmic statement:
\[ \mathbf{\log_a N = x} \]
Where:
- \(a\) is the Base (it must be positive and \(a \ne 1\)).
- \(x\) is the Index or Exponent (this is the logarithm itself).
- \(N\) is the Number (this must be positive, \(N > 0\)).
Analogy: Think of it like a mathematical switch. If you need to find the power \(x\), you use the log switch.
Quick Review of Basic Log Identities
- Since \(a^1 = a\), it follows that: \(\mathbf{\log_a a = 1}\)
- Since \(a^0 = 1\), it follows that: \(\mathbf{\log_a 1 = 0}\)
Key Takeaway: Logarithms and indices are two sides of the same coin. Learn to switch effortlessly between \(a^x = N\) and \(\log_a N = x\).
2. The Laws of Logarithms (The Rules of Operation)
These laws allow you to simplify complicated expressions and, crucially, help you solve equations where the unknown is in the index.
Let \(X\) and \(Y\) be positive numbers, and \(p\) be any real number.
Law 1: The Multiplication Law (Product Rule)
When multiplying numbers, you add their logarithms:
\[ \mathbf{\log_a (XY) = \log_a X + \log_a Y} \]
Memory Aid: "Multiplication in the number side becomes Addition on the log side."
Law 2: The Division Law (Quotient Rule)
When dividing numbers, you subtract their logarithms:
\[ \mathbf{\log_a \left(\frac{X}{Y}\right) = \log_a X - \log_a Y} \]
Memory Aid: "Division in the number side becomes Subtraction on the log side."
Law 3: The Power Law (Crucial for Solving Equations!)
The power of the number can be brought down as a multiplier:
\[ \mathbf{\log_a (X^p) = p \log_a X} \]
This is the most important rule for solving exponential equations (e.g., finding \(x\) in \(5^x = 100\)).
Common Mistake to Avoid:
DO NOT confuse \(\log_a (X+Y)\) with \(\log_a X + \log_a Y\). There is NO simplification rule for the log of a sum (or difference).
Key Takeaway: Use the power law (Law 3) whenever your variable is stuck in the exponent!
3. The Natural Exponential Function: \(e^x\) and \(\ln x\)
In A-Level Mathematics, we mostly use a special irrational base called \(e\) (Euler's number), approximately \(e \approx 2.71828\). This number is fundamental to calculus because the gradient of \(y = e^x\) is always equal to its value at any point.
The Functions
- The Natural Exponential Function: \(\mathbf{y = e^x}\)
- The Natural Logarithmic Function: \(\mathbf{y = \ln x}\)
This is the exponential function with base \(e\).
This is the logarithm to base \(e\). \(\ln x\) is shorthand for \(\log_e x\).
Relationship as Inverse Functions
Since \(\ln x\) is just a logarithm with base \(e\), the inverse rule applies:
- \(\mathbf{e^{\ln x} = x}\)
- \(\mathbf{\ln (e^x) = x}\)
Did you know? \(e\) is often called the "natural" base because it arises naturally when describing continuous growth processes, like those found in calculus.
Graphs and Properties of \(y = e^{kx}\) and \(y = \ln x\)
1. Graph of \(y = e^x\)
- Passes through \((0, 1)\) (since \(e^0 = 1\)).
- It is an increasing function.
- Domain: \(x \in \mathbb{R}\) (all real numbers).
- Range: \(y > 0\).
- The \(x\)-axis (\(y=0\)) is a horizontal asymptote.
2. Graph of \(y = \ln x\)
- Passes through \((1, 0)\) (since \(\ln 1 = 0\)).
- It is an increasing function.
- Domain: \(\mathbf{x > 0}\). (You cannot take the log of a negative number or zero!)
- Range: \(y \in \mathbb{R}\) (all real numbers).
- The \(y\)-axis (\(x=0\)) is a vertical asymptote.
The graphs of \(y = e^x\) and \(y = \ln x\) are reflections of each other in the line \(y = x\).
The Function \(\mathbf{y = e^{kx}}\)
The graph of \(y = e^{kx}\) shows accelerated growth if \(k\) is positive, and exponential decay if \(k\) is negative.
- If \(k > 0\), the function increases very quickly (growth).
- If \(k < 0\), the function decreases, approaching the \(x\)-axis (decay).
Key Takeaway: Always remember the domains and ranges! The domain restriction on \(\ln x\) (\(x>0\)) is very important when checking final answers or sketching graphs.
4. Solving Logarithmic and Exponential Equations
The main technique for solving equations where the variable is in the index is to introduce logarithms, usually \(\ln\) because it is on your calculator and works well with base \(e\).
Step-by-Step: Solving Exponential Equations
Example: Solve \(3 \times 2^{3x-1} = 5\).
Step 1: Isolate the Exponential Term
Divide by 3: \(2^{3x-1} = \frac{5}{3}\)
Step 2: Take Logarithms of Both Sides
Apply \(\ln\) (or any log base) to both sides:
\[
\ln(2^{3x-1}) = \ln\left(\frac{5}{3}\right)
\]
Step 3: Use the Power Law
Bring the index down as a multiplier:
\[ (3x - 1) \ln 2 = \ln\left(\frac{5}{3}\right) \]
Step 4: Solve for \(x\)
Expand and rearrange (treat \(\ln 2\) and \(\ln(5/3)\) as constants):
\[
3x \ln 2 - \ln 2 = \ln\left(\frac{5}{3}\right)
\]
\[
3x \ln 2 = \ln\left(\frac{5}{3}\right) + \ln 2
\]
Using the Product Law on the right side: \(\ln(5/3) + \ln 2 = \ln\left(\frac{5}{3} \times 2\right) = \ln\left(\frac{10}{3}\right)\)
\[ 3x = \frac{\ln(10/3)}{\ln 2} \] \[ x = \frac{\ln(10/3)}{3 \ln 2} \]
(Always remember to give the final numerical answer to 3 significant figures.)
Solving Logarithmic Equations
When solving log equations, the goal is often to use the laws of logarithms to combine terms, and then use the fundamental switch to convert the log statement back into an index statement.
Example: Solve \(\ln(x+2) - \ln x = 1\).
Step 1: Combine Log Terms (Quotient Law)
\[
\ln\left(\frac{x+2}{x}\right) = 1
\]
Step 2: Convert to Exponential Form
Remember, \(\ln\) is base \(e\). So \(\log_e A = 1\) becomes \(e^1 = A\):
\[
\frac{x+2}{x} = e^1
\]
Step 3: Solve for \(x\)
\[
x+2 = ex
\]
\[
2 = ex - x
\]
\[
2 = x(e - 1)
\]
\[
x = \frac{2}{e - 1} \approx 1.16
\]
Solving Inequalities in Indices
When solving an inequality like \(2^x < 5\), the process is identical to solving the equation, but you must maintain the inequality sign. Since the base (\(2\)) is greater than 1, the function is increasing, and the inequality direction does not change.
\[ 2^x < 5 \] \[ \ln(2^x) < \ln 5 \] \[ x \ln 2 < \ln 5 \] \[ x < \frac{\ln 5}{\ln 2} \]
Key Takeaway: Isolate the power, apply logarithms, use the power law, and solve algebraically. Always check your solutions, especially for log equations, to ensure they don't require taking the log of a negative number!
5. Transforming Relationships to Linear Form
A key application of logarithms in Paper 3 is simplifying a non-linear relationship (a curve) so that it becomes a linear relationship (\(Y = mX + c\)). This is often used to find unknown constants from experimental data by plotting a straight line graph.
We use logarithms to put on our "log-goggles" and see the straight line hidden within the curve.
Case 1: The Power Model \(\mathbf{y = kx^n}\)
This model relates \(y\) to a power of \(x\), where \(k\) and \(n\) are constants.
Transformation Process:
- Take natural logarithms (\(\ln\)) of both sides: \[ \ln y = \ln(kx^n) \]
- Use the Product Law (\(\ln(AB) = \ln A + \ln B\)): \[ \ln y = \ln k + \ln(x^n) \]
- Use the Power Law (\(\ln(A^n) = n \ln A\)): \[ \mathbf{\ln y = n \ln x + \ln k} \]
Linear Analysis:
This equation is now in the linear form \(Y = mX + c\), where:
- \(\mathbf{Y = \ln y}\)
- \(\mathbf{X = \ln x}\)
- Gradient \(\mathbf{m = n}\)
- Y-intercept \(\mathbf{c = \ln k}\)
If you plot a graph of \(\ln y\) against \(\ln x\), the result will be a straight line. The gradient gives you \(n\), and the intercept lets you find \(k\).
Case 2: The Exponential Model \(\mathbf{y = k(a^x)}\)
This model relates \(y\) to \(x\) in the index, where \(k\) and \(a\) are constants.
Transformation Process:
- Take natural logarithms (\(\ln\)) of both sides: \[ \ln y = \ln(k a^x) \]
- Use the Product Law: \[ \ln y = \ln k + \ln(a^x) \]
- Use the Power Law: \[ \mathbf{\ln y = (\ln a) x + \ln k} \]
Linear Analysis:
This equation is now in the linear form \(Y = mX + c\), where:
- \(\mathbf{Y = \ln y}\)
- \(\mathbf{X = x}\)
- Gradient \(\mathbf{m = \ln a}\) (The gradient is the logarithm of the base \(a\))
- Y-intercept \(\mathbf{c = \ln k}\)
If you plot a graph of \(\ln y\) against \(x\), the result will be a straight line. The gradient lets you find \(a\), and the intercept lets you find \(k\).
Quick Review Box: Linearisation
Model: \(y = kx^n\)
Linear Equation: \(\ln y = n \ln x + \ln k\)
Plot Y vs X: \(\ln y\) vs \(\ln x\)
Gradient (m): \(n\)
Intercept (c): \(\ln k\)
Model: \(y = k(a^x)\)
Linear Equation: \(\ln y = (\ln a) x + \ln k\)
Plot Y vs X: \(\ln y\) vs \(x\)
Gradient (m): \(\ln a\)
Intercept (c): \(\ln k\)
Key Takeaway: When faced with a linearization problem, the first step is always to take the logarithm of both sides. This converts the curve's relationship (multiplication/powers) into a straight line relationship (addition/multiplication).
6. Chapter Summary and Confidence Boost
You have covered the crucial theoretical foundation for exponential and logarithmic functions! Remember that these two functions are inseparable—they are inverses of each other.
The skills developed here—using the three laws, especially the power law, and converting non-linear models into linear forms—are essential for high marks in P3. Practice those transformation questions!
Don't worry if the algebra seems messy when dealing with terms like \(\ln 5\) or \(\ln a\). Just remember to treat those terms as simple constants until the final step. Keep practicing, and you'll master the log switch!