Step-by-step worked examples and graded practice questions on exponentials and logarithms — exponential graphs and growth/decay modelling, converting between index and logarithmic form, the laws of logarithms, and solving exponential and log equations. Written to the Edexcel Pure Year 1 specification and equally suitable for AQA and OCR A.
📚 Pure Year 1 (AS)✅ 15 Practice Questions🔍 6 Worked Examples⚠️ Common Mistakes
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An exponential function has the variable in the power, like \(y=2^x\) — completely different behaviour from the polynomials you've met so far. A logarithm is simply the inverse question: "what power do I need?" The two ideas are two sides of the same coin, and this topic moves fluently between them.
Five skills make up this topic:
Exponential graphs — recognising the shape, intercept and asymptote of \(y=a^x\)
Growth and decay modelling — using an exponential formula to model a real situation
The log definition — converting fluently between index form and logarithmic form
The laws of logarithms — combining or splitting log expressions
Solving equations — using logs to solve exponential equations, and using log laws to solve log equations
Exponential graphs
Every graph of \(y=a^x\) (for \(a>1\)) shares the same key features: it passes through \((0,1)\) since \(a^0=1\) for any \(a\), it never touches the \(x\)-axis (\(y=0\) is a horizontal asymptote, approached but never reached), and it increases without bound as \(x\) increases. Replacing \(x\) with \(-x\) reflects the graph in the \(y\)-axis, turning growth into decay.
Worked Example 1
Sketch the graphs of \(y=2^x\) and \(y=2^{-x}\) on the same axes, stating the \(y\)-intercept and the equation of the asymptote for each.
1
Both curves pass through \((0,1)\), since \(2^0=2^{-0}=1\)
2
\(y=2^x\) increases as \(x\) increases (growth); \(y=2^{-x}\) decreases as \(x\) increases (decay) — each is the other reflected in the \(y\)-axis
3
Neither curve ever reaches \(y=0\): both have asymptote \(y=0\)
AnswerBoth pass through \((0,1)\) with asymptote \(y=0\)
\(y=2^x\) (teal, growth) and \(y=2^{-x}\) (amber, decay) meet at the shared \(y\)-intercept \((0,1)\), marked. Both approach — but never reach — the asymptote \(y=0\).
Growth and decay modelling
A real quantity that grows or shrinks by the same proportion over each equal time interval is modelled by \(y=A b^t\), where \(A\) is the starting value and \(b\) is the growth factor per time period (\(b>1\) for growth, \(0<b<1\) for decay).
Worked Example 2
The number of bacteria in a culture is modelled by \(P=500(1.2)^t\), where \(t\) is the time in hours. Find (a) the initial number of bacteria, (b) the number after 6 hours, to the nearest whole number.
1
(a) Substitute \(t=0\): \(P=500(1.2)^0=500(1)\)
2
(b) Substitute \(t=6\): \(P=500(1.2)^6\)
3
Evaluate: \(P=500\times2.985984\)
Answer(a) 500 bacteria; (b) 1493 bacteria
The log definition
A logarithm answers "what power?" — \(\log_a x = n\) means exactly the same thing as \(a^n=x\). Being able to swap fluently between these two forms is the foundation for everything else in this topic.
\[\log_a x = n \iff a^n = x\]
Worked Example 3
Write (a) \(\log_2 32=5\) in index form, (b) \(5^3=125\) in logarithmic form.
1
(a) The base is 2, the answer is 5: \(2^5=32\)
2
(b) The base is 5, the power is 3: \(\log_5 125=3\)
Answer(a) \(2^5=32\); (b) \(\log_5 125=3\)
The laws of logarithms
Three laws let you combine or split logarithms, directly mirroring the index laws you already know:
\[\log_a(xy) \equiv \log_a x + \log_a y\]
\[\log_a\left(\frac{x}{y}\right) \equiv \log_a x - \log_a y\]
\[\log_a(x^n) \equiv n\log_a x\]
Worked Example 4
Write \(\log a + 3\log b - \log c\) as a single logarithm.
1
Use the power law on \(3\log b\): \(= \log a + \log b^3 - \log c\)
2
Use the product law: \(\log a + \log b^3 = \log(ab^3)\)
3
Use the quotient law: \(\log(ab^3) - \log c = \log\left(\dfrac{ab^3}{c}\right)\)
Answer\(\log\left(\dfrac{ab^3}{c}\right)\)
Solving exponential equations
When the unknown is in the power and the bases don't match, take logs of both sides — the power law then brings the unknown down where you can solve for it normally.
Worked Example 5
Solve \(5^x = 20\), giving your answer to 3 significant figures.
Combine every log term into a single logarithm using the laws, then use the log definition to remove the log altogether. Always check each solution is valid — a logarithm is only defined for a positive argument, so a solution that makes the original expression negative or zero must be rejected.
Worked Example 6
Solve \(\log_2 x + \log_2(x-2) = 3\).
1
Combine using the product law: \(\log_2[x(x-2)] = 3\)
2
Use the log definition: \(x(x-2) = 2^3 = 8\)
3
Rearrange and factorise: \(x^2-2x-8=0 \implies (x-4)(x+2)=0 \implies x=4\) or \(x=-2\)
4
Reject \(x=-2\): \(\log_2(x)\) requires \(x>0\), so \(x=-2\) is invalid
Answer\(x = 4\)
Exam tips
💡 Tip 1
Always check the domain when solving log equations
Every log term needs a positive argument — substitute each candidate solution back into the ORIGINAL equation (not the rearranged version) to check none of them make a logarithm undefined.
💡 Tip 2
State the y-intercept and asymptote as two separate facts
"Passes through \((0,1)\)" and "asymptote \(y=0\)" are two distinct marks in most mark schemes — write both explicitly, don't fold them into one sentence.
💡 Tip 3
Combine logs into one before removing them
You cannot use the log definition on an equation with multiple separate log terms — use the laws first to reduce everything to a single \(\log_a(\ldots)\), then convert.
💡 Tip 4
Any base works for taking logs, but be consistent
\(\log_{10}\) (written \(\log\)) and \(\ln\) both work for solving \(a^x=b\) — just use the same base on both sides of the equation throughout.
💡 Tip 5
In modelling questions, identify \(A\) and \(b\) before calculating anything
In \(y=Ab^t\), \(A\) is always the value at \(t=0\) and \(b\) is the multiplier per time period — stating what each represents in context is often a separate mark from the calculation itself.
Common mistakes
Common Mistake 1
Confusing \(\log(x+y)\) with \(\log x + \log y\)
These are not equal — there is no law for the log of a sum. Only products, quotients and powers inside a single log can be split using the laws.
Common Mistake 2
Forgetting to reject invalid solutions of a log equation
Solving the rearranged polynomial equation is not the end of the question — any root that makes an original log argument \(\leq0\) must be explicitly discarded, not just left in the final answer.
Common Mistake 3
Applying the power law to the wrong part of an expression
\(3\log b\) becomes \(\log(b^3)\) — the coefficient 3 moves to become the power of \(b\) inside the log, it does not multiply the whole logarithm's value before you've evaluated it.
Common Mistake 4
Mixing up the growth factor with the percentage change
A 20% increase per period means \(b=1.2\), not \(b=0.2\) or \(b=20\) — always convert a percentage change to a decimal multiplier first.
Common Mistake 5
Reading off the wrong sign for the y-intercept of \(y=a^{-x}\)
Both \(y=a^x\) and \(y=a^{-x}\) pass through \((0,1)\) — the reflection changes which side grows and which decays, but the y-intercept itself never moves.
Practice questions
Work through each question before checking the worked solution.
Core Skills
Q1Write \(\log_3 9=2\) in index form.Core Skills
Q2Write \(4^2=16\) in logarithmic form.Core Skills
Q3State the \(y\)-intercept and the equation of the asymptote of \(y=5^x\).Core Skills
Q4Simplify \(\log a + \log b\).Core Skills
Q5Evaluate \(\log_2 16\).Core Skills
Exam-Style
Q6The value of a car is modelled by \(V=15000(0.85)^t\). Find the value after 5 years, to the nearest £.Exam-Style
Q7Write \(2\log a - \log b\) as a single logarithm.Exam-Style
Q8Solve \(3^x = 50\), giving your answer to 3 significant figures.Exam-Style
Q9Solve \(\log_5(x+4)=2\).Exam-Style
Q10Simplify \(\log_2 32 - \log_2 4\).Exam-Style
A* Challenge
Q11Solve \(\log_3 x + \log_3(x+6) = 3\).A* Challenge
Q12Given that \(\log_a 5 = p\) and \(\log_a 2 = q\), express \(\log_a 20\) in terms of \(p\) and \(q\).A* Challenge
Q13Solve \(2^{2x} - 5(2^x) + 4 = 0\).A* Challenge
Q14Solve \(\log_2 x = \log_4(x+6)\).A* Challenge
Q15The number of people infected by a virus is modelled by \(N=20e^{0.3t}\), where \(t\) is measured in days. Find the time taken for the number infected to reach 500, giving your answer to 3 significant figures.A* Challenge
Answers — full worked solutions
Core Skills (Q1–Q5)
Q1 — \(3^2=9\)
\(3^2=9\)
Q2 — \(\log_4 16=2\)
\(\log_4 16=2\)
Q3 — \(y\)-intercept \((0,1)\), asymptote \(y=0\)
\(y\)-intercept \((0,1)\); asymptote \(y=0\)
Q4 — \(\log(ab)\)
\(\log a + \log b = \log(ab)\)
Q5 — \(4\)
1
\(2^4=16\)
\(\log_2 16 = 4\)
Exam-Style (Q6–Q10)
Q6 — £6656
1
\(V = 15000(0.85)^5 = 15000 \times 0.443705\)
\(V \approx £6656\)
Q7 — \(\log\left(\dfrac{a^2}{b}\right)\)
1
\(2\log a = \log(a^2)\)
\(\log(a^2) - \log b = \log\left(\dfrac{a^2}{b}\right)\)
Every \(y=a^x\) graph (\(a>1\)) passes through \((0,1)\) with asymptote \(y=0\); replacing \(x\) with \(-x\) reflects growth into decay.
Growth/decay modelling uses \(y=Ab^t\), where \(A\) is the starting value and \(b\) is the multiplier per time period.
\(\log_a x = n \iff a^n=x\) — the log definition converts freely between index and log form.
The three log laws (product, quotient, power) combine or split logarithms; use them to reduce a log equation to a single logarithm before converting back to index form.
Always check that a candidate solution of a log equation keeps every original log argument positive — reject any that don't.
This topic builds on the index laws from GCSE Indices and the growth/decay graph work from GCSE Non-Linear Graphs, and reuses the factorising and quadratic-solving skills from Quadratics. Next in Pure Year 1: Vectors.
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