For Integrated Programme students: Your current school materials, teacher instructions, and assessment scope take precedence because IP topic sequence and depth vary by school. This is an Eclat IP guide, not the O-Level / SEC G3 exam-track guide.
How this chapter applies
Eclat core: exponential and logarithmic functions, graphs, laws, change of base, equations, and models form the main route.
School-sensitive extension: more demanding parameter models, asymptotic analysis, and links to calculus may vary by school.
2027 national comparison: K341 Topic A6 covers exponential and logarithmic functions, their graphs, logarithm laws, change of base, simple equations, and models.
Check your school: confirm the required bases, graph sketching conventions, and modelling depth.
Q: What does IP AMaths Notes (Upper Sec, Year 3-4): 02) Logarithms and Exponentials cover? A: Log laws, change of base, and exponential modelling with inverse relationships for IP Additional Mathematics.
Logarithms invert exponentials. Switch comfortably between both forms to linearise growth models and solve equations involving powers.
Keep the full topic roadmap handy via our IP Maths tuition hub so you can jump into related drills, quizzes, or diagnostics as you move through these notes.
The core idea is simple: Logarithms undo exponentials.
Use it as a working check: Before using log laws, check the domain: every logged expression must be positive. After solving, reject any root that breaks that condition.
Then go one layer deeper: Example: if log(x + 3) + log(x - 1) appears, the domain is x greater than 1 before any quadratic solving starts.
Choosing the first transformation
Logarithm questions usually become short once you choose the right form. Start by identifying what needs to be isolated or straightened.
Question cue
First transformation
Check before accepting the answer
Several logs with the same base are added or subtracted
Combine them into one log using product or quotient laws.
Every original logged expression must be positive, not just the combined expression.
A log equals a number
Convert to index form.
Keep the base condition a>0, a=1 in mind.
The unknown is in an exponent
Take logs on both sides after isolating the exponential term.
Do not take logs of a negative or zero quantity.
The model has form y=Abx
Take logs so the graph becomes linear.
Identify which logged quantity is the vertical-axis variable.
The model uses ekx
Use natural logarithms to solve for the exponent.
Round only at the end so the time or constant is not distorted.
Common trap:log(a+b) is not loga+logb. Log laws apply to multiplication, division, and powers after the expression inside the log is already factored.
Domain Checkpoint Before Log Laws
Write the domain restriction before combining or expanding logs.
Original form
First domain condition
Why this comes first
loga(x−2)
x>2
The expression inside the log must be positive before any solving starts.
loga(x+3)+loga(x−1)
x>1
Both original log inputs must be positive, not just the combined product.
loga((x−4)2)
x=4
loga!(x−5x+2)
x−5x+2>0
Misconception check: a root that solves the final algebraic equation is not automatically valid. Test it in the original log expressions, because the original domain is the filter.
1 Core facts
Definition: logab=c iff ac=b for a>0, a=1, b>0.
Laws mirror index rules: loga(xy)=logax+logay
Power rule: loga(xk)=klogax.
Change of base: logab=logcalogcb
2 Worked example - Solve logarithmic equation
Solve log2(x+3)+log2(x−1)=3.
Combine using product law: log2((x+3)(x−1))=3.
Convert to index form: (x+3)(x−1)=23=8.
Expand: x2+2x−3=8 so x2+2x−11=0
Solve quadratic: x=2−2±4+44=2−2±48=−1±23
Domain check: require x>1, so x=−1−23
3 Worked example - Linear law
The model y=Abx can be linearised.
Given data points (x,y)=(1,6.2) and (4,20.5), show the straight-line form and estimate A and b.
When a modelling question asks for a straight-line graph, decide which quantity becomes the vertical axis before reading the gradient and intercept.
For y=Abx, plot logy against x. The gradient is logb and the intercept is logA.
For y=Aekx, plot lny against x. The gradient is k
For y=Axn, plot logy against logx. The gradient is n
Worked check: if the straight-line graph of log10y against x has gradient 0.25 and intercept 0.60, then b=100.25 and A=100.60. Do not report 0.25 and 0.60 as the constants in the original model.
Common trap: the graph intercept belongs to the transformed equation, not automatically to the original model. Undo the logarithm before writing A or b.
4 Worked example - Exponential growth time
A culture of bacteria follows N(t)=120×exp((9/50)t), where t is measured in weeks. Determine how long it takes for the population to reach 500.
Set up the equation 500=120×exp((9/50)t).
Divide by 120 to obtain exp((9/50)t)=25/6.
Take natural logarithms: (9/50)t=ln(25/6).
Solve for t: t=(50/9)×ln(25/6)≈7.93 weeks(to 3 s.f.).
Quick check: substituting t=7.93 gives N(t)≈500.1, consistent with rounding.
Answer: the culture reaches 500 cells after approximately 7.9 weeks.
5 Practice Quiz
Consolidate log laws, change-of-base manoeuvres, and exponential graph shifts before you tackle fresh modelling questions.
6 Try this
For 3log5(x−1)=log5(9x+1), solve for x and justify any rejected roots.
A square is non-negative, but the log input cannot be zero.
,
x=5
The whole fraction must be positive; numerator and denominator do not have to be positive separately.
