Solving equations with the unknown in the power Cambridge IGCSE Additional Mathematics revision
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In plain words
In 3ˣ = 20, the unknown is up in the power, where ordinary algebra cannot reach it. Logarithms bring it down. That is what they were invented for.
4 things to know
- To solve aˣ = b, take logs of both sides: x ln a = ln b, so x = ln b ÷ ln a. Logs to base 10 work just as well.
- If both sides can be written as powers of the same number, do that and put the powers equal. No logs are needed.
- To solve e^(kx) = c, take natural logs: kx = ln c.
- When there are powers on both sides with different bases, take logs, expand, and collect the terms in x.
Worked example
Solve 3ˣ = 20, giving the answer to 3 significant figures.
- Take logs: x ln 3 = ln 20.
- x = ln 20 ÷ ln 3 = 2.7268…
- x = 2.73.
Worked example
Solve 2^(x + 1) = 5ˣ.
- Take logs: (x + 1) ln 2 = x ln 5.
- Expand and collect: x ln 2 + ln 2 = x ln 5, so ln 2 = x(ln 5 − ln 2).
- x = ln 2 ÷ (ln 5 − ln 2) = 0.756.
Tips and tricks
- ln b ÷ ln a is not ln(b ÷ a). Work out the two logs separately, then divide.
- Check by putting your answer back: 3^2.73 should come out close to 20.
It lands in your notebook with its questions as flashcards.
Solving equations with the unknown in the power: 6 questions and answers
These are the quiz’s questions. Do the quiz first, then come back here for the ones that got you.
What is the solution of 2ˣ = 16?
2⁴ = 16.
Which method solves 3ˣ = 7?
The log brings the power down.
What is the exact solution of 3ˣ = 7?
x ln 3 = ln 7.
What is the solution of eˣ = 5?
The natural log undoes e to a power.
What is the solution of 10ˣ = 1000?
10³ = 1000.
What is the solution of 9ˣ = 27?
3^(2x) = 3³, so 2x = 3.
Quiz
6 questions
Tap an answer and you’ll see straight away whether it’s right, and why.
Worksheet
3 questions, 7 marks. Write your answers on paper, then check them.
Solving equations with the unknown in the power
Cambridge IGCSE Additional Mathematics 0606 · 7 marks · papermunch.org
Name ______________________________ Date ______________
Solve 5ˣ = 30, giving the answer to 3 significant figures.[2]
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x = 2.11. ln 30 ÷ ln 5.
Solve 4ˣ = 8, giving an exact answer.[2]
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x = 3/2. Both are powers of 2: 2^(2x) = 2³, so 2x = 3.
Solve e^(2x − 1) = 7, giving an exact answer.[3]
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x = (1 + ln 7)/2. 2x − 1 = ln 7.
Answers: Solving equations with the unknown in the power
- 1. x = 2.11. ln 30 ÷ ln 5.
- 2. x = 3/2. Both are powers of 2: 2^(2x) = 2³, so 2x = 3.
- 3. x = (1 + ln 7)/2. 2x − 1 = ln 7.