Secondary 4 Additional Mathematics Learning Guide | Exponential and Logarithmic Functions and Equations

Exponential and logarithmic functions are two views of the same relationship. Exponentials describe repeated multiplicative change. Logarithms reverse that process by asking which exponent produced a given value. In Secondary 4 Additional Mathematics, the strongest students do not treat these as separate chapters. They move between powers, logarithms, graphs, equations and calculus as one connected system.

This Secondary 4 Additional Mathematics Learning Guide develops exponential and logarithmic reasoning from first principles to examination control. It belongs to the Additional Mathematics Learning Hub and connects directly to functions, algebra, trigonometry, differentiation, integration and mathematical modelling.

Power → exponential function → inverse relationship → logarithm → equation → graph → model → calculus.

Exponential change is multiplicative, not additive

A linear relationship changes by equal differences. An exponential relationship changes by equal ratios. If a quantity doubles each time x increases by 1, the values may go 3, 6, 12, 24, 48. The differences are not constant, but the ratio is: each new value is twice the previous one.

Input change	Linear pattern	Exponential pattern
x increases by 1	Add the same amount	Multiply by the same factor
Typical form	y=mx+c	y=Abx
Graph behaviour	Constant gradient	Gradient usually changes with x

The general exponential form

A common exponential model is y=Ab^x. If b>1, the function grows as x increases. If 0<b<1, the function decays. The parameter A sets the starting scale because y=A when x=0.

Worked Example 1 | Read an exponential model

A quantity is modelled by P=500(1.04)^t. At t=0, P=500, so 500 is the initial value. The factor 1.04 means the quantity is multiplied by 1.04 for each one-unit increase in t, equivalent to a 4% increase per period.

Index laws are the algebra underneath exponentials

• a^maⁿ=a^m+n
• a^m/aⁿ=a^m−n
• (a^m)ⁿ=a^mn
• a⁰=1 for a≠0
• a⁻ⁿ=1/aⁿ
• a^1/n is an nth root

If these rules are insecure, logarithmic work becomes harder because every later equation depends on earlier exponent control.

Solve exponential equations by matching bases when possible

If both sides can be written with the same positive base, equate the exponents.

Worked Example 2 | Same-base equation

Solve 2^x+1=16. Since 16=2⁴, x+1=4 and x=3.

Worked Example 3 | Rewrite both sides using a common base

Solve 8^x=4^x+1. Rewrite 8=2³ and 4=2², giving 2^3x=2^2x+2. Hence 3x=2x+2 and x=2.

Logarithms are inverse exponents

The statement a^x=y is equivalent to log_ay=x. A logarithm asks: what exponent on base a produces y?

• 2⁵=32 means log₂32=5
• 10³=1000 means log₁₀1000=3
• e⁰=1 means ln 1=0

Why logarithm inputs must be positive

For a positive base a, a^x is always positive. Therefore its inverse logarithmic function only accepts positive real inputs. In ordinary real-number A-Math work, log_ax is defined for x>0.

This is a major source of invalid solutions. Algebra can produce a candidate that must still be rejected because it makes a logarithm argument zero or negative.

The three main logarithm laws

• Product: log(MN)=log M+log N
• Quotient: log(M/N)=log M−log N
• Power: log(M^k)=k log M

These are exponent laws seen through an inverse function. Multiplication of powers adds exponents; division subtracts exponents; raising a power to another power multiplies exponents.

Worked Example 4 | Expand a logarithmic expression

Expand log(x³y/√z), assuming positive variables.

3log x + log y − (1/2)log z.

Worked Example 5 | Condense logarithms

Write 2log x + log y − log z as one logarithm.

log(x²y/z).

The most common false logarithm rule

There is no rule log(M+N)=log M+log N. Logarithm laws convert multiplication into addition, not addition into addition. For example, log(10+10)=log20 while log10+log10=log100.

Common logarithms and natural logarithms

log x often denotes base 10 in school calculator work; ln x denotes logarithm to base e. The number e is approximately 2.71828 and is especially important in continuous growth, decay and calculus.

Changing logarithm base

log_ab = log b/log a = ln b/ln a.

Worked Example 6 | Change of base

log₂7 = ln7/ln2 ≈ 2.807. This is plausible because 2²=4 and 2³=8.

Solve logarithmic equations by combining first when possible

Worked Example 7 | Simple logarithmic equation

Solve log₂x=5. Convert to exponential form: x=2⁵=32.

Worked Example 8 | Solve using logarithm laws

Solve log(x−1)+log(x+1)=log15.

Domain: x−1>0 and x+1>0, so x>1. Combine: log[(x−1)(x+1)]=log15. Hence x²−1=15, so x=±4. The domain leaves only x=4.

Solve exponential equations with logarithms when bases cannot be matched

Worked Example 9 | Unknown exponent

Solve 3^x=10. Taking logs gives xlog3=log10=1, so x=1/log3≈2.096.

Worked Example 10 | Exponential model with unknown time

A quantity follows N=200(1.08)^t. Find t when N=300. Then 1.5=(1.08)^t, so t=log1.5/log1.08≈5.27.

Graphs of exponential functions

• y=a^x passes through (0,1)
• all y-values are positive
• y=0 is a horizontal asymptote
• domain is all real x
• range is y>0
• for a>1 the graph increases; for 0<a<1 it decreases

Graphs of logarithmic functions

y=log_ax is the inverse of y=a^x. Its graph passes through (1,0), has domain x>0, has all real y-values, and has vertical asymptote x=0. Exponential and logarithmic graphs reflect in y=x.

Worked Example 11 | Connect inverse coordinates

If (3,8) lies on y=2^x, then (8,3) lies on y=log₂x.

Transformations of exponential and logarithmic graphs

For y=Ae^k(x−h)+d, A changes vertical scale, k controls horizontal scale and growth or decay direction, h shifts horizontally, and d moves the horizontal asymptote to y=d. For y=log(x−h)+d, the vertical asymptote moves to x=h.

Natural exponential growth and decay

The model y=Ae^kt gives growth when k>0 and decay when k<0. Dividing by A and taking natural logs gives ln(y/A)=kt, which releases an unknown time from the exponent.

Worked Example 12 | Continuous decay

Q=800e^−0.12t. Find t when Q=400. Then 0.5=e^−0.12t, so ln0.5=−0.12t and t≈5.78.

Doubling time and half-life

For y=Ae^kt, doubling time T satisfies T=ln2/k for k>0. For decay, a half-life H satisfies H=ln(1/2)/k, with k<0.

Worked Example 13 | Growth constant from doubling time

A quantity doubles every 6 time units. Then 2=e^6k, so k=ln2/6≈0.1155.

Differentiation of exponential functions

d/dx(e^x)=e^x. For u=u(x), d/dx(e^u)=e^uu′. The chain factor remains essential.

Worked Example 14 | Differentiate an exponential function

For y=5e^2x−1, dy/dx=10e^2x−1.

Differentiation of logarithmic functions

For x>0, d/dx(lnx)=1/x. More generally, d/dx[lnu]=u′/u for a positive differentiable inner function u.

Worked Example 15 | Differentiate a logarithmic composite

For y=ln(3x+1), dy/dx=3/(3x+1). The original function requires x>−1/3.

Integration connects back through the same inverse relationships

∫e^xdx=e^x+C, while ∫1/x dx=ln|x|+C. These are calculus versions of the exponential-logarithmic inverse system.

Linearising an exponential relationship

If y=Ae^kx, taking natural logs gives ln y = ln A + kx. A plot of ln y against x is therefore linear with gradient k and intercept ln A.

Worked Example 16 | Interpret a linearised model

If ln y=0.4x+1.2, then y=e^1.2e^0.4x. The linearised gradient 0.4 is the exponential growth constant.

Equations that become quadratic after substitution

When 4^x and 2^x appear together, use 4^x=(2^x)² and let u=2^x.

Worked Example 17 | Hidden quadratic in an exponential equation

Solve 4^x−5·2^x+4=0. Let u=2^x. Then u²−5u+4=0=(u−1)(u−4), so u=1 or 4. Hence x=0 or x=2.

Quadratic structure in logarithms

If (logx)² and logx appear, let u=logx, solve the quadratic in u, then return to x.

Worked Example 18 | Quadratic in log x

Solve (logx)²−3logx+2=0. Let u=logx. Then (u−1)(u−2)=0, so logx=1 or 2. For base 10, x=10 or 100.

Domain restrictions should be written before solving

If an equation contains ln(x−2), then x>2. If it contains separate logarithms log(x−1)+log(x+3), both arguments must be positive separately. Algebraic manipulation can hide these original restrictions, so write them first.

Worked Example 19 | Logarithmic domain control

Solve ln(x−2)=ln(6−x). Domain conditions give 2<x<6. Equal logs imply x−2=6−x, so x=4, which satisfies the domain.

Use graph intersections to understand equation solutions

An equation such as e^x=3x asks where y=e^x and y=3x intersect. A graph can reveal the likely number and location of solutions and can provide a plausibility check even when algebra or numerical methods supply the final values.

Modelling: discrete percentage growth and continuous growth are not identical

P=P₀(1+r)^t represents repeated multiplication by 1+r each discrete period. P=P₀e^kt represents continuous exponential change. The parameters must be interpreted from the exact model rather than treated as interchangeable.

Verification tools

• Substitute the solution into the original equation.
• Check that every logarithm argument is positive.
• Bracket an exponent between nearby known powers.
• Confirm a growth model increases and a decay model decreases.
• Check t=0 against the stated initial value.
• Use graph intersections as a structural check.
• Attach units and contextual meaning to the final variable.

Common failure modes

Visible error	Likely cause	Repair move
log(a+b)=log a+log b	Product law overgeneralised	Test with simple numbers
Negative logarithm argument accepted	Domain not checked	Write positivity restrictions first
Logs used when bases already match	Representation not simplified	Look for common powers first
Exponent not brought down correctly	Power law insecure	Write log(ax)=xlog a
e3x differentiated as e3x	Chain factor missing	Differentiate the exponent
ln(3x+1) differentiated as 1/(3x+1)	Composite structure missed	Multiply by inner derivative 3
Growth factor 1.08 read incorrectly	Factor and percentage confused	Separate 1+0.08
Decay model gives impossible time	Signs mishandled	Inspect ln of a number below 1
Quadratic exponential structure missed	Repeated function not recognised	Substitute u=2x
Calculator result accepted without checking	Inverse meaning ignored	Compare with nearby powers

The equation decision tree

1. Is the unknown in an exponent, a logarithm, or both?
2. Can the exponential bases be rewritten to match?
3. Can logarithms be combined using valid laws?
4. What domain restrictions apply?
5. Is there a hidden quadratic structure?
6. Do logarithms need to be taken to release an exponent?
7. Does the final candidate satisfy the original equation and domain?

Examination control: exact structure first, calculator second

Preserve exact structure until the equation has been simplified. Then evaluate the final logarithmic quotient if a decimal is requested. A reliable sequence is rewrite → simplify → state domain → solve symbolically → approximate → verify.

Independent practice

1. Solve 3^x+1=81.
2. Solve 9^x=27.
3. Evaluate log₂64.
4. Expand log(x²y³/z).
5. Write 3log a−2log b as one logarithm.
6. Solve log x=2.5.
7. Solve log(x−2)+log x=log8.
8. Solve 5^x=18 to 3 d.p.
9. Solve 4^x−6·2^x+8=0.
10. P=1000(1.06)^t. Find t when P=1500.
11. Q=500e^−0.2t. Find t when Q=100.
12. Differentiate 7e^4x.
13. Differentiate ln(5x−2).
14. Integrate 3e^3x.
15. Integrate 4/x.

Explained answers

1. x=3. 2. x=3/2. 3. 6. 4. 2logx+3logy−logz. 5. log(a³/b²). 6. x=10^2.5≈316.228. 7. x=4 after domain checking. 8. x=log18/log5≈1.796. 9. x=1 or 2. 10. t≈6.96. 11. t≈8.05. 12. 28e^4x. 13. 5/(5x−2). 14. e^3x+C. 15. 4ln|x|+C.

Teaching sequence: build the inverse system before the laws

Begin with multiplicative patterns and exponential graphs. Stabilise index laws. Introduce logarithms as the inverse question “what exponent gives this output?” Build logarithm laws from exponent laws, then move into equations with domain restrictions. Connect graphs as inverses, apply growth and decay models, and finally mix the topic with differentiation, integration, functions and algebra so the student recognises the structure when the chapter label disappears.

How this guide connects to the rest of Secondary 4 A-Math

• Differentiation, tangents, normals and stationary points
• Integration, definite integrals and area under curves
• Trigonometric functions, identities and equations
• How functions connect tables, graphs and equations

Final thought

Exponentials and logarithms become much easier when the student sees one reversible relationship underneath the notation. Exponential notation asks what a power produces. Logarithmic notation asks what power produced it.

Recognise the base. Protect the domain. Release the exponent. Return the answer to the model.

Return to the Additional Mathematics Learning Hub.