Logarithms and Their Laws – OCR FSMQ Additional Maths

EL2 — $\log_a x$ as the inverse of $a^x$

$\log_a x = y \;\Longleftrightarrow\; a^y = x$

Say it in words: $\log_a x$ is the power you must raise $a$ to in order to get $x$. Almost every logarithm question can be answered by translating it into that sentence.

Logarithm form	Index form	Read as
$\log_2 8 = 3$	$2^3 = 8$	"$2$ to the power $3$ is $8$"
$\log_{10} 1000 = 3$	$10^3 = 1000$	"$10$ cubed is $1000$"
$\log_5 25 = 2$	$5^2 = 25$	"$5$ squared is $25$"
$\log_3 \tfrac19 = -2$	$3^{-2} = \tfrac19$	"$3$ to the power $-2$ is $\tfrac19$"
$\log_7 1 = 0$	$7^0 = 1$	Anything to the power $0$ is $1$

Notation conventions

$\log x$ with no base written means base $10$
$\ln x$ means base $\mathrm{e}$ — the natural logarithm

The inverse relationships

$\log_a\left(a^x\right) = x$ · $a^{\log_a x} = x$
Each operation completely undoes the other.

Worked Example 1 — Evaluating without a calculator

Evaluate (a) $\log_4 64$, (b) $\log_2 \tfrac{1}{16}$, (c) $\log_9 3$.

①(a) What power of $4$ gives $64$? $4^3 = 64$, so the answer is $3$.

②(b) $2^4 = 16$, so $2^{-4} = \tfrac{1}{16}$. The answer is $-4$.

③(c) What power of $9$ gives $3$? Since $9^{1/2} = 3$, the answer is $\tfrac12$.

Fractional and negative answers are perfectly normal. A logarithm can be any real number — it is the input that must be positive, not the output.

The input must be positive. $\log_a x$ requires $x > 0$, because $a^y$ is always positive. There is no logarithm of $0$ or of a negative number.

The laws, as the specification lists them

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

Where they come from. Each law is an index law in disguise. Because $a^p \times a^q = a^{p+q}$, multiplying numbers corresponds to adding their logarithms. The other two laws follow from the division and power index laws in the same way.

Worked Example 2 — Combining into one logarithm

Write $2\log 3 + \log 5 - \log 9$ as a single logarithm.

①Deal with the coefficient first: $2\log 3 = \log\left(3^2\right) = \log 9$

②$\log 9 + \log 5 = \log 45$

③$\log 45 - \log 9 = \log\!\left(\dfrac{45}{9}\right) = \log 5$

Always apply the power law first. Coefficients must be absorbed into the logarithm before you can combine anything.

Worked Example 3 — Splitting one up

Express $\log\!\left(\dfrac{x^3 y}{z^2}\right)$ in terms of $\log x$, $\log y$ and $\log z$.

①The quotient law: $\log\left(x^3 y\right) - \log\left(z^2\right)$

②The product law on the first part: $\log\left(x^3\right) + \log y - \log\left(z^2\right)$

③The power law: $3\log x + \log y - 2\log z$

Worked Example 4 — Solving a logarithmic equation

Solve $\log_2 x + \log_2 (x - 2) = 3$.

①Combine the left side: $\log_2\big[x(x-2)\big] = 3$

②Convert to index form: $x(x-2) = 2^3 = 8$

③$x^2 - 2x - 8 = 0$, so $(x-4)(x+2) = 0$.

④$x = 4$ or $x = -2$.

⑤Reject $x = -2$: it would require $\log_2(-2)$, which does not exist.

⑥$x = 4$. Check: $\log_2 4 + \log_2 2 = 2 + 1 = 3$ ✓

Always check for rejected solutions. Combining logarithms can introduce answers that make the original expression undefined. Stating the rejection and the reason earns a mark.

Tempting but wrong	Why, and what is actually true
$\log(x+y) = \log x + \log y$	No. $\log x + \log y = \log(xy)$. There is no law for $\log$ of a sum.
$\dfrac{\log x}{\log y} = \log\!\left(\dfrac{x}{y}\right)$	No. The quotient law needs the division inside: $\log x - \log y$.
$(\log x)^2 = 2\log x$	No. $2\log x = \log\left(x^2\right)$, which is different from $(\log x)^2$.
$\log(xy) = \log x \times \log y$	No. Multiplying inside gives adding outside.

Worked Example 5 — Testing a false claim with numbers

Disprove the claim that $\log(x+y) = \log x + \log y$.

①Take $x = y = 10$.

②Left side: $\log(10+10) = \log 20 = 1.301$

③Right side: $\log 10 + \log 10 = 1 + 1 = 2$

④$1.301 \neq 2$, so the claim is false.

A single numerical counter-example settles it. If you are ever unsure whether a "law" is real, test it with $x = y = 10$ in base $10$ — the arithmetic is trivial.

Features of $y = \log_a x$ for $a > 1$

Passes through $(1, 0)$ · defined only for $x > 0$
Increasing, but ever more slowly · the $y$-axis is a vertical asymptote

The mirror-image relationship. Because logarithms and exponentials are inverses, their graphs are reflections in $y = x$. Every feature swaps over: the exponential's $y$-intercept $(0,1)$ becomes the logarithm's $x$-intercept $(1,0)$, and the horizontal asymptote becomes a vertical one.

Definition

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

In words

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

$\log$ alone

Means base $10$.

Product law

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

Quotient law

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

Power law

$n\log x = \log\left(x^n\right)$ — use it first.

$\log_a 1 = 0$

And $\log_a a = 1$.

No law for

$\log$ of a sum.

Domain

$x > 0$ only.

Check answers

Reject any making a logarithm undefined.

Question 1

Write $\log_5 125 = 3$ in index form.

▶ Show solution

$5^3 = 125$

Question 2

Evaluate $\log_2 32$, $\log_{10} 100$ and $\log_6 1$.

▶ Show solution

$2^5 = 32$, so $\log_2 32 = 5$.

$10^2 = 100$, so $\log_{10} 100 = 2$.

$6^0 = 1$, so $\log_6 1 = 0$.

Question 3

Evaluate $\log_3 \tfrac{1}{27}$.

▶ Show solution

$3^3 = 27$, so $3^{-3} = \tfrac{1}{27}$.

$\log_3 \tfrac{1}{27} = -3$

Question 4

Evaluate $\log_{16} 4$.

▶ Show solution

$16^{1/2} = 4$, so $\log_{16} 4 = \dfrac{1}{2}$.

Question 5

Write $\log 8 + \log 5 - \log 2$ as a single logarithm.

▶ Show solution

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

$\log 40 - \log 2 = \log 20$

Question 6

Write $3\log 2 - 2\log 4$ as a single logarithm, and evaluate it.

▶ Show solution

$3\log 2 = \log 8$ and $2\log 4 = \log 16$.

$\log 8 - \log 16 = \log\!\left(\dfrac{8}{16}\right) = \log 0.5$

$= -0.301$ (3 s.f.)

Question 7

Express $\log\!\left(\dfrac{a^2}{b}\right)$ in terms of $\log a$ and $\log b$.

▶ Show solution

$\log\left(a^2\right) - \log b = 2\log a - \log b$

Question 8

Solve $\log_3 x = 4$.

▶ Show solution

$x = 3^4 = 81$

Question 9

Solve $\log_5 x + \log_5 4 = 2$.

▶ Show solution

$\log_5 (4x) = 2$

$4x = 5^2 = 25$

$x = 6.25$

Check: $\log_5 6.25 + \log_5 4 = \log_5 25 = 2$ ✓

Question 10

(a) Solve $\log_2 (x+3) + \log_2 (x-1) = 5$. (b) Explain why one solution must be rejected. (c) Solve $\log x + \log(x-3) = 1$. (d) Show that $\log_a b \times \log_b a = 1$ for any valid bases.

▶ Show solution

(a) Combine: $\log_2\big[(x+3)(x-1)\big] = 5$

$(x+3)(x-1) = 2^5 = 32$

$x^2 + 2x - 3 = 32$, so $x^2 + 2x - 35 = 0$.

$(x+7)(x-5) = 0$, giving $x = -7$ or $x = 5$.

(b) $x = -7$ makes $x + 3 = -4$ and $x - 1 = -8$, so the original expression would need $\log_2(-4)$ and $\log_2(-8)$ — neither exists.

Only $x = 5$ is valid. Check: $\log_2 8 + \log_2 4 = 3 + 2 = 5$ ✓

The spurious root appears because combining the logarithms produced $(x+3)(x-1)$, which is positive when both brackets are negative — a case the original equation never allowed.

(c) $\log\big[x(x-3)\big] = 1$, and base $10$ gives

$x(x-3) = 10^1 = 10$

$x^2 - 3x - 10 = 0$, so $(x-5)(x+2) = 0$.

$x = 5$ or $x = -2$; reject $x = -2$ as $\log(-2)$ is undefined.

$x = 5$. Check: $\log 5 + \log 2 = \log 10 = 1$ ✓

(d) Let $\log_a b = p$, so by the definition $a^p = b$.

Taking $\log_b$ of both sides: $\log_b\left(a^p\right) = \log_b b = 1$.

By the power law, $p\log_b a = 1$.

Substituting back $p = \log_a b$:

$\log_a b \times \log_b a = 1$ as required.

Verify with numbers: $\log_2 8 = 3$ and $\log_8 2 = \tfrac13$, and $3 \times \tfrac13 = 1$ ✓