Logarithmic and exponential functions in O-Level Additional Mathematics (4049): the laws of logarithms, the graphs of exponential and logarithmic functions, solving exponential and logarithmic equations, and the linear law

Two inverse operations

Logarithms and exponentials are two views of the same relationship, and this Algebra-strand topic of Additional Mathematics (SEAB 4049) is built on the fact that they undo each other. An exponential function raises a base to a power; a logarithm asks "what power gives this number?" Holding that inverse relationship in mind makes the laws, the graphs and the equation-solving all hang together rather than feeling like separate rules.

This module covers four outcomes: the laws of logarithms, the graphs of the two function families, solving exponential and logarithmic equations, and the linear law for transforming a non-linear relationship into a straight line. This overview ties the four dot points together; each has its own worked answers and practice.

The laws of logarithms

The laws of logarithms outcome gives the algebra of logarithms: the product law $\log_a(xy) = \log_a x + \log_a y$, the quotient law $\log_a\left(\dfrac{x}{y}\right) = \log_a x - \log_a y$, and the power law $\log_a(x^n) = n\log_a x$. With $\log_a 1 = 0$, $\log_a a = 1$ and the change-of-base formula $\log_a x = \dfrac{\log_b x}{\log_b a}$, you can simplify and evaluate logarithmic expressions.

Graphs of the two families

The graphs of exponential and logarithmic functions outcome covers their shapes and the inverse relationship. The exponential $y = a^x$ passes through $(0, 1)$ with the $x$-axis as a horizontal asymptote, while the logarithm $y = \log_a x$ passes through $(1, 0)$ with the $y$-axis as a vertical asymptote. Because they are inverses, each is the reflection of the other in the line $y = x$.

Solving exponential and logarithmic equations

The exponential and logarithmic equations outcome uses the inverse relationship to solve. For an unknown in the exponent you take logarithms and apply the power law; for an unknown inside a logarithm you convert to index form. Crucially, you must reject any solution that makes the argument of a logarithm zero or negative.

The linear law

The linear law outcome transforms a non-linear law into $Y = mX + c$ so its constants can be read from a straight-line graph. For $y = ax^n$, taking logarithms gives $\lg y = n\lg x + \lg a$, so a plot of $\lg y$ against $\lg x$ has gradient $n$ and intercept $\lg a$. For $y = ab^x$, $\lg y = (\lg b)x + \lg a$ is linear in $x$.

How this module is examined

• Both papers, all questions. Paper 1 and Paper 2 (each 2 hours 15 minutes, 90 marks, 50 percent) cover the full syllabus, and you answer every question.
• Apply the laws in the right direction. Combine logarithms into one before converting to index form; expand them when you need to isolate a term.
• Identify $Y$, $X$, $m$ and $c$ explicitly. In a linear-law question, state what you are plotting and which combination of constants the gradient and intercept give before reading values.

Check your knowledge

Short questions across the four outcomes. Work them with full method, then check the solutions.

1. Write $\log_a 12$ in terms of $\log_a 2$ and $\log_a 3$. (2 marks)
2. Solve $3^x = 20$, giving your answer to 3 significant figures. (2 marks)
3. Solve $\log_5(2x - 1) = 2$. (2 marks)
4. The law $y = ax^n$ is to be tested by plotting $\lg y$ against $\lg x$. State the gradient and the vertical intercept of the resulting line in terms of $a$ and $n$. (2 marks)