Unit 1: Pure Mathematics — Logarithms
Welcome to the study notes for Logarithms! If you have ever wondered how to find an unknown power, like solving for \(x\) in \(2^x = 10\), logarithms are the exact tool you need. In this unit, we will unlock the relationship between powers and logarithms, master three simple laws, solve equations with powers, and transform curves into straight lines. Don't worry if this seems brand new at first — we will break down every idea step by step.
1. What is a Logarithm? (Definition & Equivalence)
At its heart, a logarithm is simply the inverse operation of a power (index). A logarithm answers the question: "To what power must I raise the base to get this number?"
The Fundamental Equivalence Rule
Every index statement can be rewritten as an equivalent logarithmic statement:
\(y = b^x \iff \log_b(y) = x\)
Where:
• \(b\) is the base (the number being raised to a power).
• \(x\) is the index / power / exponent.
• \(y\) is the result / argument.
Let's look at familiar examples:
• \(2^3 = 8 \iff \log_2(8) = 3\) ("To what power must base 2 be raised to get 8? The answer is 3.")
• \(10^2 = 100 \iff \log_{10}(100) = 2\)
• \(5^4 = 625 \iff \log_5(625) = 4\)
• \(3^{-2} = \frac{1}{9} \iff \log_3\left(\frac{1}{9}\right) = -2\)
Two Special Rules to Remember
No matter what base \(b\) you use (as long as \(b > 0\) and \(b \neq 1\)):
1. \(\log_b(1) = 0\) (because \(b^0 = 1\))
2. \(\log_b(b) = 1\) (because \(b^1 = b\))
Standard Base 10 Convention
In GCSE Further Mathematics, when a logarithm is written without a base subscript (such as \(\log x\)), it almost always refers to base 10 (\(\log_{10} x\)). This is the standard "log" button on your scientific calculator!
Key Takeaway: The expression \(\log_b(y) = x\) means exactly the same thing as \(b^x = y\). Keep the base at the bottom!
2. The Three Basic Laws of Logarithms
Just like indices have rules (e.g. \(a^m \times a^n = a^{m+n}\)), logarithms have their own matching set of three laws. These laws only apply when all terms have the same base.
Law 1: The Multiplication Law (Product Rule)
\(\log(ab) = \log a + \log b\)
Adding two logs with the same base allows you to multiply their contents inside a single log.
• Example: \(\log_{10}(2) + \log_{10}(5) = \log_{10}(2 \times 5) = \log_{10}(10) = 1\)
Law 2: The Division Law (Quotient Rule)
\(\log\left(\frac{a}{b}\right) = \log a - \log b\)
Subtracting two logs with the same base allows you to divide their contents inside a single log.
• Example: \(\log_2(24) - \log_2(3) = \log_2\left(\frac{24}{3}\right) = \log_2(8) = 3\)
Law 3: The Power Law (Index Rule)
\(\log(a^n) = n\log a\)
A power inside a logarithm can "jump down" to become a multiplier at the front.
• Example: \(\log_{10}(1000) = \log_{10}(10^3) = 3\log_{10}(10) = 3(1) = 3\)
• Example: \(\log(x^5) = 5\log x\)
Worked Example: Writing as a Single Logarithm
Question: Simplify \(2\log a + 3\log b - \log c\) into a single logarithm.
Step 1: Apply the power law to bring multipliers inside as powers:
\(\log(a^2) + \log(b^3) - \log c\)
Step 2: Apply the multiplication law to the addition terms:
\(\log(a^2 b^3) - \log c\)
Step 3: Apply the division law to the subtraction term:
\(\log\left(\frac{a^2 b^3}{c}\right)\)
Note: The CCEA GCSE specification focuses on these three laws. Change of base is outside this specification scope.
Key Takeaway: Addition outside becomes multiplication inside; subtraction outside becomes division inside; powers inside become multipliers outside.
3. Solving Indicial (Index) Equations
An indicial equation is an equation where the unknown variable is in the power (index), such as \(4^x = 19\). You cannot solve this using basic algebra alone, but by taking logarithms of both sides (using base 10), we can bring the unknown power down using the Power Law!
Type 1: Single Linear Power (\(a^x = b\))
Question: Solve \(3^x = 20\), giving your answer to 3 significant figures.
Step 1: Take logs (base 10) of both sides:
\(\log(3^x) = \log(20)\)
Step 2: Use the Power Law to bring the power \(x\) to the front:
\(x\log(3) = \log(20)\)
Step 3: Divide by \(\log(3)\) and calculate:
\(x = \frac{\log(20)}{\log(3)}\)
\(x = \frac{1.30103}{0.47712} \approx 2.73\) (to 3 s.f.)
Type 2: Equations with Powers on Both Sides (\(a^{mx+c} = b^{kx+d}\))
Question: Solve \(5^{2x+1} = 7^{x-3}\), giving your answer to 3 decimal places.
Step 1: Take logs of both sides:
\(\log(5^{2x+1}) = \log(7^{x-3})\)
Step 2: Use the Power Law. Always use brackets around the index!
\((2x + 1)\log 5 = (x - 3)\log 7\)
Step 3: Expand both sets of brackets:
\(2x\log 5 + \log 5 = x\log 7 - 3\log 7\)
Step 4: Collect all terms containing \(x\) on one side and constant terms on the other:
\(2x\log 5 - x\log 7 = -3\log 7 - \log 5\)
Step 5: Factorise out \(x\):
\(x(2\log 5 - \log 7) = -(3\log 7 + \log 5)\)
Step 6: Rearrange and evaluate using your calculator:
\(x = \frac{-(3\log 7 + \log 5)}{2\log 5 - \log 7}\)
\(x = \frac{-(3 \times 0.84510 + 0.69897)}{2(0.69897) - 0.84510} = \frac{-3.23427}{0.55284} \approx -5.850\) (to 3 d.p.)
Key Takeaway: Taking logs of both sides frees the unknown from the power. When expanding brackets, remember that \(\log 5\) and \(\log 7\) are simply constant numbers.
4. Logarithmic Graphs & Reduction to Linear Form
In science and mathematics, many relationships produce curved lines (such as \(y = ax^n\) or \(y = Ab^x\)). By taking logarithms of both sides, we can transform these curves into the standard straight-line equation:
\(Y = mX + c\)
where \(m\) is the gradient and \(c\) is the vertical intercept.
Model 1: The Power Relationship (\(y = ax^n\))
• Original formula: \(y = ax^n\)
• Take logs of both sides: \(\log y = \log(ax^n)\)
• Apply the Multiplication Law: \(\log y = \log a + \log(x^n)\)
• Apply the Power Law: \(\log y = \log a + n\log x\)
• Rearrange to match \(Y = mX + c\):
\(\log y = n(\log x) + \log a\)
If we plot a graph of \(\log y\) on the vertical axis against \(\log x\) on the horizontal axis:
• The gradient (\(m\)) \(= n\)
• The vertical intercept (\(c\)) \(= \log a \implies a = 10^c\)
Model 2: The Exponential Relationship (\(y = Ab^x\) or \(y = ka^x\))
• Original formula: \(y = Ab^x\)
• Take logs of both sides: \(\log y = \log(Ab^x)\)
• Apply the Multiplication Law: \(\log y = \log A + \log(b^x)\)
• Apply the Power Law: \(\log y = \log A + x\log b\)
• Rearrange to match \(Y = mX + c\):
\(\log y = (\log b)x + \log A\)
If we plot a graph of \(\log y\) on the vertical axis against \(x\) on the horizontal axis:
• The gradient (\(m\)) \(= \log b \implies b = 10^m\)
• The vertical intercept (\(c\)) \(= \log A \implies A = 10^c\)
Quick Comparison Table:
• Power Model (\(y = ax^n\)): Plot \(\log y\) vs \(\log x\) \(\longrightarrow\) Gradient \(= n\), Intercept \(= \log a\)
• Exponential Model (\(y = Ab^x\)): Plot \(\log y\) vs \(x\) \(\longrightarrow\) Gradient \(= \log b\), Intercept \(= \log A\)
Worked Example: Finding Constants from a Graph
Question: Variables \(x\) and \(y\) are related by the equation \(y = ax^n\). A graph of \(\log y\) against \(\log x\) produces a straight line with a gradient of \(2.5\) and a vertical intercept of \(0.602\). Find the values of \(a\) and \(n\).
Step 1: State the linear form: \(\log y = n\log x + \log a\)
Step 2: Match the gradient: \(n = \text{gradient} = 2.5\)
Step 3: Match the intercept: \(\log a = 0.602\)
Step 4: Find \(a\) using the inverse log: \(a = 10^{0.602} \approx 4\)
Conclusion: The formula is \(y = 4x^{2.5}\).
Key Takeaway: Look closely at the axes! If the horizontal axis is \(\log x\), it is a power law (\(y = ax^n\)). If the horizontal axis is plain \(x\), it is an exponential law (\(y = Ab^x\)). To find the constant from the intercept, always calculate \(10^{\text{intercept}}\).
5. Pitfalls and Common Examiner Mistakes
CCEA examiners frequently highlight the following common mistakes. Make sure to avoid them:
1. Inventing Fake "Distributive" Laws:
• Wrong: \(\log(a + b) = \log a + \log b\)
• Right: \(\log(a \times b) = \log a + \log b\)
Remember: Logarithms cannot be multiplied across a plus sign!
2. Confusing Log Division with the Quotient Rule:
• Wrong: \(\frac{\log a}{\log b} = \log(a - b)\) or \(\log\left(\frac{a}{b}\right) = \frac{\log a}{\log b}\)
• Right: \(\log\left(\frac{a}{b}\right) = \log a - \log b\)
3. Forgetting Brackets in Indicial Equations:
• When taking logs of \(3^{2x+1}\), writing \(2x + 1\log 3\) is incorrect because only the 1 is multiplied by \(\log 3\).
• Correct: \((2x + 1)\log 3\).
4. Forgetting to Undo Logs on Graph Intercepts:
• The vertical intercept gives \(\log a\), not \(a\) itself.
• If the intercept is \(c = 1.3\), then \(a = 10^{1.3} \approx 19.95\), NOT \(a = 1.3\).
5. Graph Axis Confusion:
• Always check whether the horizontal axis is labelled \(x\) or \(\log x\) before deciding which formula to use!
Quick Summary Checklist
Before sitting your Unit 1 exam, make sure you can confidently:
• Convert between index form \(y = b^x\) and log form \(\log_b(y) = x\).
• Recall and apply all 3 log laws: \(\log(ab)\), \(\log(a/b)\), and \(\log(a^n)\).
• Solve indicial equations by taking logs (base 10) of both sides and using brackets.
• Convert \(y = ax^n\) and \(y = Ab^x\) into straight-line form \(Y = mX + c\).
• Determine constants from gradients and intercepts using powers of 10.