DSE Maths | Logarithms and Exponential Equations Tutor Prep Guide | Log Laws, Taking Logs and Four Common Errors

2026-11-05

Tutor lesson-prep guides · Secondary Maths

Students can recite that a log is the inverse of a power, but faced with 2 to the power of 3x minus 1 equals 32 they test values one by one, or split the log of a sum into a sum of logs. This lesson turns the equations into linear ones and drills one universal move: take logs on both sides.

Objectives — Students can (1) split log 32 into log 8 plus log 4 (5 equals 5) and explain why the log of a sum is not the sum of logs; (2) solve 2 to the power of 3x minus 1 equals 32 by taking logs, giving 3x minus 1 equals 5, so x equals 2; (3) use P equals 8000 times 2 to the power of t over 6 to show that 18 years is 3 doubling periods and 64000, and reverse the same model to find the 5.42 years needed for a value to halve at a factor of 0.88 per year.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Index laws	What is a to the m times a to the n?	a to the m plus n, powers add for the same base
Log and index conversion	What is log base 2 of 8?	3, because 2 cubed is 8
Calculator use	What base do the log and ln keys use?	log is base 10, ln is base e, and log 32 over log 2 can be keyed in

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-6 min	Prior knowledge check	Ask index laws, log definition, calculator keys	Answer orally, errors go in the corner
6-22 min	Act one: log laws	Derive log of a product from 2 cubed times 2 squared	Verify all three laws themselves
22-38 min	Act two: taking logs	Model the full solution of 2 to the 3x minus 1 equals 32	Follow with two questions
38-50 min	Act three: growth and decay	From a doubling time of six years to eighteen, then reverse	Write the equation with the unknown in the index
50-56 min	Question drill	One of each of four types, six minutes	Independent, then swap and mark
56-60 min	Exit check	Collect three answers and go through them	Books closed

03 Teaching script (read it out)

Scene 1: Act one: the log laws are the index laws read backwards

Write the index law first: same base multiplied means powers add. Say the same fact in log language and it becomes log of a product equals sum of logs. Check with numbers: 2 cubed is 8 and 2 squared is 4, and their product is 32. On the right log 8 plus log 4 is 3 plus 2, which is 5, and log 32 is also 5.
Give all three laws together: product becomes sum, quotient becomes difference, and a power moves to the front. The third one matters most because it moves the unknown out of the index and into a coefficient. Make them use it immediately: log 9 equals 2 log 3, because 9 is 3 squared.
Then the trap: the log of a sum is not the sum of logs. log of 8 plus 4 is log 12, while log 8 plus log 4 is log 32, and the two are not equal. Have them key both into a calculator and see 1.079 against 1.505. One calculator check beats ten verbal warnings.
Finish with change of base: log base a of b equals log b over log a, so log base 2 of 32 can be computed as log 32 over log 2. That single formula lets every answer be checked numerically. Misconception box: logs do not distribute over addition, so log of a plus b must never be split.

Students restate: product to sum, quotient to difference, power to the front, and brackets cannot be split.

Scene 2: Act two: when the unknown sits in the index, take logs on both sides

Write 2 to the power of 3x minus 1 equals 32. Students will try x equals 1, 2, 3 and lose time. The correct move is to take the log of both sides. Since 32 is 2 to the power of 5, the indices must match, so 3x minus 1 equals 5 and x equals 2. Substitute back: 2 to the fifth is 32, which checks.
Now a case with a non-integer answer: 3 to the x equals 20. Taking logs gives x log 3 equals log 20, so x equals log 20 over log 3, about 2.73. Stress that answers need not be integers; a decimal is the normal case, not a mistake.
Third variation, different bases: 2 to the x equals 5 to the x minus 1. After taking logs, collect the x terms: x log 2 equals (x minus 1) log 5. Expanding and gathering x is the step that decides every harder question, so students must do it themselves at least once.
Misconception box: taking the log of one side only breaks the equation. Both sides must be treated the same way.

Students restate: take logs on both sides, bring powers down, gather the x terms.

Scene 3: Act three: growth and decay share one formula with different bases

Start with doubling: a culture doubles every 6 hours and starts at 8000. 18 hours is 3 doubling periods, giving 8000 times 2 cubed, which is 64000. Have them write the model P equals P0 times 2 to the power of t over T and explain that t over T counts the periods.
Then reverse it with a value that is not a whole number of periods: when does the count reach 20000? Solve 2 to the power of t over 6 equals 2.5 and get about 7.97 hours. That step is where act two becomes useful.
Decay uses the same form with a base below one: a machine worth 20000 loses 12 percent a year, so it is multiplied by 0.88 each year, and after five years it is worth 10555. To find when it halves, solve 0.88 to the power of t equals 0.5, giving t equals ln 0.5 over ln 0.88, about 5.42 years. A compound interest comparison uses the same method: 5000 growing at 4 percent reaches 6000 in about 4.65 years, only the base changes to 1.04.
Close with a question: why can we not just divide by 12 percent? Because percentage decay multiplies by a shrinking base each year. The answer separates students who understand exponential change from those who memorised a formula.

Students restate: multiply for growth, take logs to reverse, base above one grows and below one decays.

04 Question types and worked steps

Type	Example	Steps	Answer
Simplify with log laws	Write log 32 as a sum of two logs	1 factorise 32, 2 apply product law, 3 check numerically	log 8 plus log 4, that is 3 plus 2 equals 5
Solve an exponential equation	Solve 2 to the 3x minus 1 equals 32	1 write both sides to the same base, 2 equate indices, 3 solve	3x minus 1 equals 5, x equals 2
Different bases	Solve 3 to the x equals 20	1 take logs, 2 bring power down, 3 divide by the coefficient	x equals log 20 over log 3, about 2.73
Growth and decay	Value 20000 multiplied by 0.88 a year, when does it halve	1 write the equation, 2 take logs, 3 solve and give units	about 5.42 years

Sanity check (teach this): Sanity check: a growth answer must exceed the start value (64000 against 8000) and a decay answer must be below it (10555 against 20000); a wrong direction means the base is wrong. The number of doubling periods is dimensionless (3); writing three years confuses periods with time. Check the base before judging a log value: log 32 is 1.505 in base 10 but 5 in base 2. A depreciation half-life must be longer than the naive one over the rate; two years would mean the value falls by a fixed 0.12 rather than times 0.88.

05 Independent practice (3 questions)

Question	Answer	What to watch
Simplify log 8 plus log 4 and verify with log 32.	log 32, that is 5	Used the product law, not bracket splitting
Solve 2 to the 3x minus 1 equals 32.	x equals 2	Took logs of both sides to the same base
A machine worth 20000 is multiplied by 0.88 each year; find the value after five years and the years to halve.	10555 after five years, about 5.42 years to halve	Wrote an equation with the unknown in the index

06 Common mistakes and what to say

Mistake	Student writes	Say this
Splitting a log of a sum	Writes log of 8 plus 4 as log 8 plus log 4	Logs have no distributive law; key both in and you will see 1.079 is not 1.505.
Taking the log of one side	Logs the left side only	Both sides must take the same base, otherwise the equality is gone.
Dividing for growth	Multiplies 8000 by 3	Three doubling periods multiply by 8, not by 3.
Treating decay as a fixed amount	Subtracts 12 percent of 20000 every year	Percentage decay multiplies by 0.88 each year, so the amount removed shrinks.

07 Homework (paper, section and question numbers)

Redo this chapter — All three practice questions (1-4), showing each log step
Maths subject index — Open the indices and logarithms chapter and do question 2, checking with a calculator
Build a comparison table — Place growth, decay and compound interest side by side with one real example and units each

08 Exit check (last 5 minutes)

Question	Answer	Decision
What is log 8 plus log 4	log 32, that is 5	Answer log 12 means repeat act one
Solve 2 to the 3x minus 1 equals 32	2	Wrong means repeat act two
Multiplied by 0.88 a year, how long to halve	about 5.42 years	Answer four years means repeat act three

09 Tutor FAQ

Why must we take logs?

Logs move the unknown out of the index into a coefficient, leaving a linear equation.

Does log or ln matter?

Any base works as long as both sides use it; the answer is identical.

Can we just divide by the growth rate?

Only as a very short-term approximation; over years the exponential model is required.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.