Use logarithms and exponentials: apply the laws of logarithms, convert between exponential and logarithmic form, and solve equations of the form a to the power x equals b.

What this dot point is asking

A logarithm answers the question "what power do I raise the base to?", and CCEA GCSE Further Mathematics uses logarithms to undo exponentials. You must move fluently between index form and logarithmic form, apply the three laws of logarithms to simplify expressions, and solve equations where the unknown is in the power, such as $a^x = b$. Logarithms are the inverse of exponential functions, so this topic builds directly on indices and functions.

Logarithm and index form

A logarithm and a power say the same thing in two ways. The statement $a^x = b$ in index form becomes $\log_a b = x$ in logarithmic form. Reading this fluently both ways is the heart of the topic.

For example, $2^5 = 32$ is the same fact as $\log_2 32 = 5$, and $\log_{10} 1000 = 3$ because $10^3 = 1000$. The base of the logarithm is the base of the power; the value inside the log is the result of the power; and the logarithm itself is the exponent.

The laws of logarithms

The log laws follow directly from the index laws, because logarithms are exponents. They let you combine or split logarithms to simplify expressions and to solve equations.

The power law is the most important for solving equations, because it moves an unknown exponent into a multiplying position where it can be isolated. Combining the laws lets you condense an expression to a single logarithm, or expand one for analysis.

Solving exponential equations

When the unknown is in the exponent, ordinary algebra cannot reach it, but logarithms can.

Exponential growth and decay

Exponential expressions model quantities that multiply by a fixed factor each step, such as compound growth or radioactive decay. A model like $P = P_0 \times r^t$ has a starting value $P_0$ and a multiplier $r$ each period: when $r > 1$ the quantity grows, and when $0 < r < 1$ it decays. To find the time $t$ for the quantity to reach a target, you rearrange to isolate the power, then take logarithms and use the power law, exactly as above. This is where logarithms become a practical tool rather than a pure-algebra exercise, and CCEA often sets such applied contexts.

For example, suppose a population is modelled by $P = 500 \times 1.08^t$ and you need the year in which it first exceeds $1000$. Setting $500 \times 1.08^t = 1000$ gives $1.08^t = 2$. Taking logs, $t\log 1.08 = \log 2$, so $t = \dfrac{\log 2}{\log 1.08} = 9.0$ to one decimal place. Because $t$ must be a whole number of years for the population to have exceeded the target, the answer is the $10$th year. Rounding up rather than down here is the kind of interpretation step the exam rewards, because nine full years still leaves the population just below $1000$.

Why this matters

Logarithms complete the picture begun by indices and functions: they are the inverse function that lets you solve for an unknown power, which no amount of ordinary rearranging can do. They appear in growth and decay problems and connect to the graph work, since exponential and logarithmic graphs are reflections of each other in $y = x$. Confident use of the log laws keeps these questions short and accurate.