IB · Mathematics: Analysis & Approaches SL & HL · Number & algebra

Exponents & logarithms: IB Mathematics: Analysis & Approaches questions with answers

Free Exponents & logarithms questions for IB Mathematics: Analysis & Approaches (SL & HL), part of Number & algebra. Each question has the answer and the marking points an examiner would look for.

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Exponents & logarithms practice questions

Question 1Structured · 6 marksA* challenge

Solve the equation 2^{2x + 2} − 17(2^x) + 4 = 0.

[6]

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Part	Answer and marking points	Marks
	x = 2, x = −2 2^{2x + 2} = 4(2^x)² (M1) 4u² − 17u + 4 = 0, where u = 2^x A1 (u − 4)(4u − 1) = 0 (M1) u = 4 or u = 14 A1 x = 2 A1 x = −2 A1	6

Question 2Structured · 5 marks

Solve the equation 4^x + (2^x) − 6 = 0, giving your answers in exact form.

[5]

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Part	Answer and marking points	Marks
	x = 1 only let u = 2^x: u² + u − 6 = 0 (M1) (u − 2)(u + 3) = 0 A1 2^x = 2 (A1) 2^x = −3 is impossible since 2^x > 0 for all x R1 x = 1 A1	5

Question 3Structured · 6 marksA* challenge

Solve the equation 2^{2x + 1} − 17(2^x) + 8 = 0.

[6]

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Part	Answer and marking points	Marks
	x = 3, x = −1 2^{2x + 1} = 2(2^x)² (M1) 2u² − 17u + 8 = 0, where u = 2^x A1 (u − 8)(2u − 1) = 0 (M1) u = 8 or u = 12 A1 x = 3 A1 x = −1 A1	6

Question 4Structured · 6 marksA* challenge

Solve the equation 2log₉(x + 4) − log₃(x − 3) = 1.

[6]

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Part	Answer and marking points	Marks
	x = 132 change of base: log₉(x + 4) = log₃(x + 4)log₃9 M1 = 12log₃(x + 4), so the equation becomes log₃(x + 4) − log₃(x − 3) = 1 A1 log₃x + 4x − 3 = 1 (M1) x + 4 = 3(x − 3) A1 x = 132 A1 check: x − 3 = 72 > 0, so the solution is valid R1	6

Question 5Structured · 5 marks

Solve the equation 9^x − 5(3^x) − 36 = 0, giving your answers in exact form.

[5]

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Part	Answer and marking points	Marks
	x = 2 only let u = 3^x: u² − 5u − 36 = 0 (M1) (u − 9)(u + 4) = 0 A1 3^x = 9 (A1) 3^x = −4 is impossible since 3^x > 0 for all x R1 x = 2 A1	5

Question 6Structured · 6 marks

Solve the equation log₃x + log₃(x − 24) = 4.

[6]

Show answers and mark scheme

Part	Answer and marking points	Marks
	x = 27 log₃(x(x − 24)) = 4 M1 x(x − 24) = 3⁴ (M1) x² − 24x − 81 = 0 A1 (x − 27)(x + 3) = 0 (A1) x = −3 is rejected because log₃ of a negative number is not defined R1 x = 27 A1	6

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Original IB-style practice questions written for mentorriyaz.in. These are not IB past paper questions, and this site is not affiliated with or endorsed by the International Baccalaureate Organization. Updated 2026-11-04. Paper generator · mentorriyaz.in