Exponentials and logarithms: A-level Maths OCR revision notes

1.06

Logarithms and exponential equations

Logarithms as inverse exponentials 1.06

Definitions

Logarithm: $lo g_{a} x = y$ means exactly $a^{y} = x$ , where the base satisfies $a > 0$ and $a \neq = 1$ , and the argument satisfies $x > 0$ .
Natural logarithm: $ln x$ is shorthand for $lo g_{e} x$ , the logarithm to base $e$ ; an unsubscripted $lo g x$ on a calculator means $lo g_{10} x$ .

Key results

A logarithm is an exponent: $lo g_{a} x$ is the power to which $a$ must be raised to give $x$ .
$lo g_{a} a = 1$ and $lo g_{a} 1 = 0$ , for every valid base $a$ .
$a^{l o g_{a} x} = x$ and $lo g_{a} (a^{x}) = x$ — the two statements that $lo g_{a}$ and $a^{x}$ undo one another.

Notes

Only positive arguments have a logarithm, so $lo g_{a} x$ is undefined for $x \leq 0$ ; every solution of a logarithmic equation must therefore be checked against the original equation before it is accepted.
The graph of $y = lo g_{a} x$ is the reflection of $y = a^{x}$ in the line $y = x$ : it passes through $(1, 0)$ , has the $y$ -axis as a vertical asymptote, and is defined only for $x > 0$ .
The base restrictions $a > 0$ , $a \neq = 1$ exist for the same reason a negative or unit base makes $a^{y}$ either ambiguous or constant — $lo g_{a} x$ inherits every restriction on the exponential it inverts.

Laws of logarithms 1.06

Key results

Product law: $lo g_{a} (x y) = lo g_{a} x + lo g_{a} y$ .
Quotient law: $lo g_{a} (\frac{x}{y}) = lo g_{a} x - lo g_{a} y$ .
Power law: $lo g_{a} (x^{k}) = k lo g_{a} x$ , for any real $k$ .
Reciprocal case of the power law: $lo g_{a} (\frac{1}{x}) = - lo g_{a} x$ .
Change of base: $lo g_{a} x = \frac{ln x}{ln a}$ , obtained by taking natural logarithms of $a^{l o g_{a} x} = x$ .

Notes

Each law mirrors an index law exactly, because that is where it comes from: multiplying powers adds their exponents, and a logarithm is an exponent, so $lo g_{a} (x y) = lo g_{a} x + lo g_{a} y$ .
There is no law for $lo g_{a} (x + y)$ ; it does not simplify at all, and writing $lo g_{a} (x + y) = lo g_{a} x + lo g_{a} y$ is one of the most heavily penalised errors in this topic.
Likewise $(lo g_{a} x)^{2}$ and $lo g_{a} (x^{2})$ are different objects — only the second one equals $2 lo g_{a} x$ .
The change-of-base formula works with any base on the right, not just $ln$ — $lo g_{2} 10 = \frac{lo g _{10} 10}{lo g _{10} 2}$ is just as valid, useful when a calculator only has a base- $10$ key.

Solving equations with an unknown exponent 1.06

Key results

$a^{x} = b ⟺ x = \frac{ln b}{ln a}$ , for $a > 0$ , $a \neq = 1$ and $b > 0$ .
When the unknown appears in exponents on both sides, as in $2^{x + 1} = 5^{2 x}$ , the same method gives $(x + 1) ln 2 = 2 x ln 5$ and hence $x = \frac{ln 2}{2 ln 5 - ln 2}$ .

Method

Take logarithms of both sides of the equation, using $ln$ or $lo g_{10}$ consistently throughout.
Use the power law to bring each unknown exponent down as a multiplier.
Collect the terms in the unknown on one side and factorise it out.
Divide to make the unknown the subject, then evaluate to three significant figures unless an exact answer is requested.

Notes

The base of the logarithm taken makes no difference to the final answer, provided the same base is used on both sides — $ln$ is the usual choice.
Taking logarithms converts a multiplicative relationship in the exponent into an additive one that ordinary linear algebra can then isolate; that is the whole reason logarithms exist.
$ln b$ where $0 < b < 1$ is negative — carry that sign carefully through the rest of the algebra rather than treating $ln b$ as automatically positive.

Equations that reduce to a quadratic 1.06

Method

Spot the repeated block: an equation in $e^{2 x}$ and $e^{x}$ is a quadratic in $e^{x}$ , since $e^{2 x} = (e^{x})^{2}$ ; an equation in $(lo g_{a} x)^{2}$ and $lo g_{a} x$ is a quadratic in $lo g_{a} x$ .
Substitute a single letter for that block, for example $u = e^{x}$ or $u = lo g_{a} x$ .
Solve the resulting quadratic in $u$ by factorising or the formula.
Reverse the substitution for each root, and discard any root that is not attainable.

Notes

$e^{x} > 0$ for every real $x$ , so a negative or zero value of $u = e^{x}$ must be rejected — it yields no solution rather than an extra one.
By contrast $lo g_{a} x$ can take any real value, so both roots of a quadratic in $lo g_{a} x$ normally survive; it is the resulting $x$ -values that must satisfy $x > 0$ .
After reversing the substitution, always re-check each candidate $x$ -value in the ORIGINAL equation, not just the substituted one — a valid $u$ can still lead to an $x$ that fails a domain condition elsewhere in the equation.

Worked examples

Worked example

Write $2 lo g_{3} 6 - lo g_{3} 4$ as a single logarithm, and hence evaluate it exactly.

Show worked solution

By the power law,

2 lo g_{3} 6 = lo g_{3} (6^{2}) = lo g_{3} 36

By the quotient law,

lo g_{3} 36 - lo g_{3} 4 = lo g_{3} (\frac{36}{4}) = lo g_{3} 9

Since $3^{2} = 9$ , the value is $2$ .

Worked example

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

Show worked solution

Take logarithms of both sides: $ln (3^{x}) = ln 20$ , so $x ln 3 = ln 20$ .

Then:

x = \frac{ln 20}{ln 3} = \frac{2.996}{1.099} = 2.73

(3 s.f.).

Worked example

Solve:

lo g_{2} x + lo g_{2} (x - 2) = 3

Show worked solution

Combine the left side with the product law:

lo g_{2} (x (x - 2)) = 3

Undo the logarithm:

x (x - 2) = 2^{3} = 8

so $x^{2} - 2 x - 8 = 0$ , which factorises as:

(x - 4) (x + 2) = 0

giving $x = 4$ or $x = - 2$ .

The value $x = - 2$ makes both original logarithms undefined, so it is rejected.

Hence $x = 4$ .

Check:

lo g_{2} 4 + lo g_{2} 2 = 2 + 1 = 3

Worked example

Solve:

e^{2 x} - 5 e^{x} + 6 = 0

giving your answers to three significant figures.

Show worked solution

Since $e^{2 x} = (e^{x})^{2}$ , put $u = e^{x}$ to get $u^{2} - 5 u + 6 = 0$ , so:

(u - 2) (u - 3) = 0

and $u = 2$ or $u = 3$ .

Both are positive, so both are attainable values of $e^{x}$ .

Taking natural logarithms:

x = ln 2 = 0.693

(3 s.f.) or $x = ln 3 = 1.10$ (3 s.f.).

1.06

The functions eˣ and ln x

The number e and the natural logarithm 1.06

Definitions

$e \approx 2.71828$ is defined so that the gradient of $y = e^{x}$ at any point equals the $y$ -value at that point — the unique base for which this is true.

Key results

$\frac{d}{d x} (e^{x}) = e^{x}$ , which is exactly why exponential models in calculus are written in terms of $e$ rather than any other base.
$e^{l n x} = x$ for $x > 0$ , and $ln (e^{x}) = x$ for all real $x$ .
$ln e = 1$ , $ln 1 = 0$ , and $ln x \to - \infty$ as $x \to 0^{+}$ .
Any exponential can be rewritten to base $e$ : $a^{x} = e^{x l n a}$ .

Notes

The graph of $y = e^{x}$ passes through $(0, 1)$ , is positive and increasing everywhere, and has the $x$ -axis as a horizontal asymptote as $x \to - \infty$ .
The graph of $y = ln x$ is its reflection in $y = x$ : domain $x > 0$ , crossing the $x$ -axis at $(1, 0)$ , increasing but ever more slowly.
$e$ is irrational, like $π$ — never round it early in a calculation; keep it as $e$ symbolically until the final numerical answer is required.

1.06

Exponential growth and decay

Exponential growth and decay models 1.06

Definitions

Exponential model: $y = A e^{k t}$ , where $A$ is the initial value of $y$ at $t = 0$ and $k$ is the growth constant.

Key results

$k > 0$ gives growth and $k < 0$ gives decay; the larger $∣ k ∣$ is, the faster the change.
Differentiating the model gives $\frac{d y}{d t} = k y$ : the rate of change is always proportional to the current amount, which is the defining feature of exponential behaviour.
The model is therefore the natural choice whenever a quantity's rate of change depends on how large the quantity currently is.
Equivalent form: $y = A b^{t}$ describes the same family, with $b = e^{k}$ ; a decay model has $0 < b < 1$ .

Notes

$k$ carries units of 'per unit time', so its numerical value depends on whether $t$ is measured in seconds, years or any other unit — always state the unit alongside it.
A pure exponential decay model never reaches zero; a model with a non-zero limiting value, such as cooling towards room temperature, takes the form $y = L + A e^{k t}$ with $k < 0$ .
$A$ is read directly from the model at $t = 0$ (since $e^{0} = 1$ ), so it never needs solving for separately if the initial value is already known or given.

Doubling time and half-life 1.06

Key results

Doubling time $T$ : $2 = e^{k T}$ , so $k = \frac{ln 2}{T}$ .
Half-life $T$ in the decay model $y = A e^{- k t}$ : $\frac{1}{2} = e^{- k T}$ , so $k = \frac{ln 2}{T}$ again.
Time to reach a general multiple $m$ of the initial value: $t = \frac{ln m}{k}$ .

Method

Write the model $y = A e^{k t}$ and substitute the stated multiple of the initial value, for example $y = 2 A$ for a doubling or $y = \frac{A}{2}$ for a half-life.
Cancel $A$ from both sides, leaving an equation in $e^{k T}$ alone.
Take natural logarithms of both sides.
Rearrange for whichever of $k$ or $T$ is unknown.

Notes

Because $A$ cancels, doubling time and half-life do not depend on the starting amount at all — a striking property of exponential change that no other simple model shares.
Be careful with signs: in $y = A e^{- k t}$ the positive constant is $k$ , so a half-life gives $k = + \frac{ln 2}{T}$ , whereas in $y = A e^{k t}$ the same decay gives $k = - \frac{ln 2}{T}$ .
After several half-lives $n$ , the remaining amount is $A (\frac{1}{2})^{n}$ — a quick mental-arithmetic route to an approximate answer without needing the exponential form at all.

Worked example

A radioactive isotope decays according to $N = N_{0} e^{- k t}$ .

Given that the isotope has a half-life of $10$ years, find the value of $k$ .

Show worked solution

After one half-life,

N = \frac{N _{0}}{2}

when $t = 10$ :

\frac{N _{0}}{2} = N_{0} e^{- 10 k}

so:

\frac{1}{2} = e^{- 10 k}

Taking natural logs:

ln (\frac{1}{2}) = - 10 k

so $- ln 2 = - 10 k$ , giving:

k = \frac{ln 2}{10} = 0.0693

(3 s.f.).

1.06

Modelling with exponentials

Linearising a model with logarithms 1.06

Key results

Exponential model: $y = A e^{k x}$ gives $ln y = ln A + k x$ , a straight line when $ln y$ is plotted against $x$ , with gradient $k$ and vertical intercept $ln A$ .
Base- $b$ exponential model: $y = A b^{x}$ gives $lo g y = lo g A + x lo g b$ , again a straight line of $lo g y$ against $x$ , with gradient $lo g b$ .
Power model: $y = A x^{n}$ gives $lo g y = lo g A + n lo g x$ , a straight line when $lo g y$ is plotted against $lo g x$ , with gradient $n$ and intercept $lo g A$ .

Method

Decide which model is proposed, and hence whether to plot $ln y$ against $x$ (exponential) or $lo g y$ against $lo g x$ (power).
Take the logarithms of the data values and plot the transformed points.
Read the gradient $m$ and intercept $c$ off the line of best fit.
Recover the parameters by reversing the logarithm, for example $A = e^{c}$ and $k = m$ for the exponential model.

Notes

The shape of the transformed plot is itself the evidence: a straight log-linear plot supports an exponential law, while a straight log-log plot supports a power law.
The intercept is a logarithm of the constant, not the constant itself — forgetting to exponentiate $c$ before quoting $A$ is the standard error here.
Match the base of logarithm used in plotting to the base used when recovering the constants — using $ln$ to plot but $lo g_{10}$ to recover $A$ (or vice versa) silently gives the wrong constant.

Worked example

The mass $m$ grams of a chemical remaining after $t$ minutes is modelled by $m = A e^{k t}$ .

Plotting $ln m$ against $t$ gives a straight line of gradient $- 0.28$ passing through the vertical axis at $3.4$ .

Find $A$ and $k$ , and find the time at which the mass has fallen to $5$ g.

Show worked solution

Taking logarithms of the model gives $ln m = ln A + k t$ , so the plotted line is:

ln m = 3.4 - 0.28 t

Comparing: $k = - 0.28$ per minute, and $ln A = 3.4$ , so:

A = e^{3.4} = 30.0

(3 s.f.).

For $m = 5$ :

ln 5 = 3.4 - 0.28 t

so:

0.28 t = 3.4 - ln 5 = 3.4 - 1.6094 = 1.7906

giving $t = 6.39$ minutes (3 s.f.).

Check:

29.96 e^{- 0.28 \times 6.3949} = 29.96 \times 0.1669 = 5.00

as required.