Exponentials and Logarithms – OCR FSMQ Additional Maths

In every function you have met so far, the variable sat in the base: $x^2$, $x^3$, $\sqrt{x}$. An exponential function puts the variable in the index instead: $2^x$, $10^x$, $1.05^x$. That single change produces growth of an entirely different character.

The contrast worth remembering. $x^{10}$ grows fast at first but $2^x$ overtakes it and never looks back. Exponential growth always wins in the end, which is why it describes compound interest, population, and the spread of anything that multiplies.

The central relationship

A logarithm is the inverse of an exponential:
$\log_a x = y \;\Longleftrightarrow\; a^y = x$

In words: $\log_a x$ is the power you must raise $a$ to in order to get $x$. That sentence is worth learning verbatim — almost every logarithm question yields to it.

Why logarithms matter here. An equation like $3^x = 20$ cannot be solved by any algebra you have met: you cannot "undo" a variable index by rooting or dividing. Logarithms are the tool that brings the $x$ down where you can reach it.

Word	Meaning
Exponential function	$y = ka^x$ — the variable is in the index.
Base	The $a$ in $a^x$ or in $\log_a x$.
Logarithm	The inverse of raising to a power.
$\log$	Written without a base, this conventionally means base $10$.
$\ln$	The natural logarithm, base $\mathrm{e}$.
Growth factor	The multiplier per period: $1.05$ for $5\%$ growth.
Decay factor	A multiplier between $0$ and $1$.
Asymptote	A line the curve approaches but never reaches.
Linear form	$Y = mX + c$ — what a log transformation produces.

You cannot take the logarithm of zero or of a negative number. There is no power of $10$ that gives $-5$, since $10^x$ is always positive. If a calculation produces $\log(-3)$, an earlier step is wrong.

EL3 — the three laws of logarithms

$\log x + \log y = \log(xy)$
$\log x - \log y = \log\!\left(\dfrac{x}{y}\right)$
$n\log x = \log\left(x^n\right)$

Two values to know

$\log_a 1 = 0$ (because $a^0 = 1$) · $\log_a a = 1$ (because $a^1 = a$)

The single most useful law

$\log\left(x^n\right) = n\log x$
This is what brings an unknown index down to ground level.

Worked Example — Solving an exponential equation

Solve $3^x = 20$, giving your answer to $3$ significant figures.

①Take logarithms of both sides: $\log\left(3^x\right) = \log 20$

②Use the power law to bring $x$ down: $x\log 3 = \log 20$

③$x = \dfrac{\log 20}{\log 3} = \dfrac{1.30103}{0.47712}$

④$x = 2.73$ (3 s.f.)

⑤Check: $3^{2.73} = 20.07 \approx 20$ ✓

$\dfrac{\log 20}{\log 3}$ is not $\log\!\left(\tfrac{20}{3}\right)$. The quotient law applies to a logarithm of a quotient, not to a quotient of logarithms. This is the single commonest error in the topic.

EL1Exponential FunctionsThe function $ka^x$ and its graph, for positive $a$.
EL2
EL3Logarithms and Their Laws$\log_a x$ as the inverse of $a^x$, and the three laws.
EL6Solving Exponential EquationsEquations of the form $a^x = b$, and disguised quadratics.
EL4
EL5Reduction to Linear FormTurning $y = ka^x$ and $y = kx^n$ into straight lines, then estimating $k$, $a$ and $n$.
EL7Growth and DecayCompound interest, populations, half-life and cooling.

$\dfrac{\log a}{\log b} \neq \log\!\left(\dfrac{a}{b}\right)$. Dividing two logarithms is not the logarithm of a quotient.

$\log(x+y) \neq \log x + \log y$. The sum law works the other way: $\log x + \log y = \log(xy)$. There is no law at all for the logarithm of a sum.

$2^x$ is not $2x$. At $x = 10$, one is $1024$ and the other is $20$.

Confusing the growth factor with the rate. $5\%$ growth means multiplying by $1.05$, not by $0.05$.

Drawing an exponential curve touching the $x$-axis. It approaches the axis asymptotically but never reaches it.

Rounding too early. $\log 20$ and $\log 3$ should be kept at full precision until the final division.

Definition

$\log_a x = y \iff a^y = x$.

In words

The power $a$ needs to give $x$.

Product

$\log x + \log y = \log(xy)$.

Quotient

$\log x - \log y = \log\!\left(\tfrac{x}{y}\right)$.

Power

$\log\left(x^n\right) = n\log x$.

$\log_a 1$

Always $0$.

Solving $a^x=b$

$x = \dfrac{\log b}{\log a}$.

Graph of $ka^x$

Through $(0,k)$, asymptote $y = 0$.

Growth factor

$1 + \dfrac{\text{rate}}{100}$.

No logs of

Zero or negative numbers.

Question 1

Write $\log_2 8 = 3$ in index form.

▶ Show solution

$2^3 = 8$

Question 2

Evaluate $\log_3 81$ and $\log_5 1$.

▶ Show solution

$3^4 = 81$, so $\log_3 81 = 4$.

$5^0 = 1$, so $\log_5 1 = 0$.

Question 3

Write $\log 6 + \log 4$ as a single logarithm.

▶ Show solution

$\log 6 + \log 4 = \log(6 \times 4) = \log 24$

Question 4

Solve $2^x = 15$, to $3$ significant figures.

▶ Show solution

$x\log 2 = \log 15$

$x = \dfrac{\log 15}{\log 2} = \dfrac{1.17609}{0.30103} = 3.91$ (3 s.f.)

Question 5

Write down the $y$-intercept of $y = 7 \times 3^x$.

▶ Show solution

At $x = 0$: $y = 7 \times 1 = 7$, so the intercept is $(0, 7)$.

Question 6

A population grows by $8\%$ a year from $5000$. Write a formula for the population after $t$ years.

▶ Show solution

Growth factor $1.08$, so $P = 5000 \times 1.08^t$.

Question 7

Simplify $3\log 2 + \log 5$.

▶ Show solution

$3\log 2 = \log\left(2^3\right) = \log 8$

$\log 8 + \log 5 = \log 40$

Question 8

Solve $\log_2 x = 5$.

▶ Show solution

By the definition, $x = 2^5 = 32$.

Question 9

Explain why $\log(-4)$ has no value.

▶ Show solution

$\log(-4)$ would be the power of $10$ giving $-4$.

But $10^x$ is positive for every real $x$ — it approaches zero for large negative $x$ but never becomes negative.

So no such power exists, and $\log(-4)$ is undefined.

Question 10

A radioactive sample of mass $80$ mg decays so that $12\%$ is lost each year.

(a) Write a formula for the mass $m$ after $t$ years. (b) Find the mass after $5$ years. (c) Find the half-life, to $3$ significant figures. (d) Explain why the sample never completely disappears according to this model, and comment on whether that is realistic.

▶ Show solution

(a) Losing $12\%$ leaves $88\%$, so the decay factor is $0.88$:

$m = 80 \times 0.88^t$

(b) $m = 80 \times 0.88^5 = 80 \times 0.527732 = 42.2$ mg (3 s.f.)

(c) Half-life is when $m = 40$:

$80 \times 0.88^t = 40 \;\Rightarrow\; 0.88^t = 0.5$

$t\log 0.88 = \log 0.5$

$t = \dfrac{\log 0.5}{\log 0.88} = \dfrac{-0.30103}{-0.055517} = 5.42$ years (3 s.f.)

Check: $0.88^{5.42} = 0.500$ ✓

(d) $0.88^t$ is positive for every $t$, however large — each year removes $12\%$ of what remains, never the last of it. So the model gives a mass that shrinks towards zero without ever reaching it: the $t$-axis is an asymptote.

Physically this is only an approximation. A real sample consists of a whole number of atoms, so once very few remain the smooth exponential model breaks down, and eventually the last atom decays and the sample genuinely is gone. The model is excellent while the number of atoms is large, and unrealistic at the very end — a limitation worth stating in any answer about long-term behaviour.