Maths › Exponentials and logarithms › Logarithms and their laws

Logarithms and their laws

A logarithm answers one question. What power produced this number? That single idea undoes every exponential, drags unknowns down out of exponents where algebra cannot otherwise reach them, and obeys three laws that turn multiplication into addition.

Year 12EDEXCEL 9MA0 6.3, 6.4, 6.5

Builds on Exponential functions and e.

IN THIS TOPIC

  • Convert between exponential and logarithmic statements, in any base and in base e.
  • Solve eax+b = p and ln (ax + b) = q by applying the right inverse to the whole side.
  • Use the three log laws to combine, split and simplify expressions.
  • Solve ax = b, and anything else with the unknown in an exponent.

COMMON MISCONCEPTION

log (x + y) = log x + log y.

The inverse of an exponential

The statement an = x and the statement loga x = n carry the same information. A logarithm is an exponent, read backwards. So log10 1000 = 3 because 103 = 1000, and loga a = 1 for every base the definition allows. Those bases are a > 0 with a ≠ 1, and the number x has to be positive as well, because an is positive for every n and no power of a will ever produce zero or a negative. The base that matters most is e, whose logarithm gets a name of its own, ln x, and it undoes last lesson's star function.

x=an⇔n=logaxx = a^{n} \iff n = \text{log}_{a} xNOT IN THE BOOKLET — LEARN IT
The curves of e to the x and its inverse, the natural logarithm, drawn at equal scales: each is the other reflected in the line y equals xy = e to the xy = ln xy = xeach is the other, reflected
FIG. 1e to the x and ln x, reflections in the line y = x. One curve passes through (0, 1), its mirror through (1, 0), and each undoes the other.

WORKED EXAMPLE

Undoing e, twice over

Solve e2x+1 = 10, and then ln (3x + 2) = 2, each to 4 significant figures.

For the first, take ln of both sides. 2x + 1 = ln 10, so x = (ln 10 − 1)/2 = 0.6513.

For the second, apply e to both sides. 3x + 2 = e2, so x = (e2 − 2)/3 = 1.796.

One of those needed the log peeled off and the other needed it applied. Working out which of the pair is in front of you is most of the skill, and the rest is rearrangement you have done since GCSE.

The three laws

Logs are exponents, so the index laws translate straight into log laws. All three are on the must-learn list,

logax+logay=logaxy\text{log}_{a} x + \text{log}_{a} y = \text{log}_{a} xyNOT IN THE BOOKLET — LEARN IT
logax−logay=logaxy\text{log}_{a} x − \text{log}_{a} y = \text{log}_{a} \frac{x}{y}NOT IN THE BOOKLET — LEARN IT
klogax=logaxkk \text{log}_{a} x = \text{log}_{a} x^{k}NOT IN THE BOOKLET — LEARN IT

and the third holds for negative and fractional k as well, so −log x = log (1/x) and ½ log x = log √x. A logarithm counts multiplicative steps, and counting is an additive business. That is the whole reason the laws look as they do. Every x and y in the three laws has to be positive, which is why the solutions of a log equation get checked back in the original, with any value that makes an argument zero or negative rejected.

Now notice what is missing from the list. Nothing on it touches log (x + y), because no law does. One line in base 10 settles it. log 20 = 1.301, while log 10 + log 10 = 2.

The powers of 2 from 1 to 32 spaced equally along a line: each doubling is one equal step, and the base-2 logarithm simply counts the steps10214283164325value: each step doubles itlog₂ of the value: each step adds 1a logarithm is a step counter for multiplication
FIG. 2Why the laws work: a log counts doublings, so multiplying values means adding step-counts. The first law in one picture.

WORKED EXAMPLE

Collapsing to a single logarithm

Write 2 log 3 + log 5 − log 15 as a single logarithm.

Move the multiple inside first. 2 log 3 = log 9.

Then combine left to right. log 9 + log 5 = log 45, and log 45 − log 15 = log (45/15).

The expression is log 3.

It is worth checking numerically as well. 0.9542 + 0.6990 − 1.1761 = 0.4771, which is log 3, and the whole check takes ten seconds on a calculator you already have in your hand.

Unknowns in the exponent

When the unknown sits in an exponent and the bases do not match, take logs of both sides. The third law then pulls the unknown down to the front as a multiplier, and ordinary algebra finishes the job.

WORKED EXAMPLE

A mismatched base

Solve 23x−1 = 3, giving the answer to 4 significant figures.

Take logs of both sides. (3x − 1) log 2 = log 3.

So 3x − 1 = log 3/log 2 = 1.585, and x = 2.585/3 = 0.8617.

Any base of logarithm will do here, log or ln alike, because the base cancels out of the ratio. The one wrong move is hunting for a way to write 3 as a power of 2 by inspection.

GUIDED PRACTICE

Solve for the exponent

Solve 5x = 30, to 4 significant figures, before opening the working.

Show the working

Take logs. x log 5 = log 30, so x = log 30/log 5 = 2.113.

A sense check brackets the answer before you write it down. 52 = 25 and 53 = 125, so x had to sit just above 2.

INDEPENDENT PRACTICE

Logs against a growth model

Last lesson's population model was P = 500e0.2t. Find, to 3 significant figures, the time at which P reaches 10 000.

Show the working

Set 500e0.2t = 10 000, so e0.2t = 20.

Take ln of both sides. 0.2t = ln 20, so t = 5 ln 20 = 15.0 days.

Every “when does the model reach…” question from here on is answered this way. The exponential poses the question and its inverse retrieves the time.

ASSESSMENT FOCUS

  • Translate between the two forms before doing any algebra. Writing an = x as loga x = n earns the setup mark on its own.
  • Name the log law at each step. Logs combined or split without a named law lose method marks even when the answer is right.
  • There is no law for log (x + y). If an addition turns up inside a log, stop and hunt for a factorisation instead.
  • For ax = b, take logs in whichever base you like and divide. Keep the exact ratio on the page until the final rounding.
  • ln undoes e and e undoes ln, but only when applied to a whole side of an equation. Never term by term.

CHECK YOURSELF

Express log 4 + log 25 as a single logarithm and evaluate it, then solve e3x = 40 to 3 significant figures.

Show a hint

4 × 25 is friendly; then ln both sides.

Show the answer

log 4 + log 25 = log 100 = 2, in base 10.

Taking ln of e3x = 40 gives 3x = ln 40, so x = ln 40/3 = 1.23.

Both answers came out of one idea. A logarithm is the exponent that was used, recovered.

A logarithm is an exponent: a to the n equals x and log base a of x equals n are one statement.

Logs turn multiplication into addition and bring exponents down as multipliers.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

Question workbook

PDF · 30 KB · print edition · white background

Question workbook

PDF · 30 KB · screen edition · navy

Answer book

PDF · 35 KB · fully worked · navy
7 questions on this topicAnswer them one at a time and mark yourself against the worked answer.Practise this topic

Or read them with their worked answers on the logarithms and their laws questions page.

CHECK YOUR PROGRESS

Rate how confident you feel with each objective for this lesson. Ratings are saved in this browser, on this device, unless you sign in.

  • Convert between exponential and logarithmic statements, in any base and in base e.
  • Solve eax+b = p and ln (ax + b) = q by applying the right inverse to the whole side.
  • Use the three log laws to combine, split and simplify expressions.
  • Solve ax = b, and anything else with the unknown in an exponent.

Open the full revision checklist to see every objective in the course in one place.