A-level · Exponentials & logarithms

Solving exponential and logarithmic equationsExam-style questions

Taking logs to solve a^x = b, equations with e^x and ln, hidden quadratics in e^x or a^x, and log equations solved with the laws — always checking which roots are valid.

14 questions · A-level level · mark scheme and worked solution on each · no sign-up to try

DifficultyAllLevel 2Level 3Level 4
  1. 1
    Solve the equation

    32x+1=5x+23^{2x+1} = 5^{x+2},

    giving your answer in the form x=ln⁡aln⁡bx = \dfrac{\ln a}{\ln b}, where aa and bb are rationals, and then to 3 significant figures. (5 marks)
    Level 3 · 5 marksTry it
  2. 2
    Solve the equation ln⁡(2x+3)=2ln⁡x\ln(2x+3) = 2\ln x. (4 marks)
    Level 3 · 4 marksTry it
  3. 3
    Solve the equation

    2log⁡3x−log⁡3(x+6)=12\log_3 x - \log_3 (x + 6) = 1,

    explaining why one root of your quadratic must be rejected. (5 marks)
    Level 3 · 5 marksTry it
  4. 4
    Solve the equation

    32x+1−10(3x)+3=03^{2x + 1} - 10\left(3^{x}\right) + 3 = 0

    giving your answers in exact form. (4 marks)
    Level 3 · 4 marksTry it
  5. 5
    (a) Show that the equation

    2log⁡3(x+1)−log⁡3(x+5)=22\log_3(x + 1) - \log_3(x + 5) = 2

    can be written as

    x2−7x−44=0x^2 - 7x - 44 = 0

    (3 marks)
    (b) (i) Solve the equation x2−7x−44=0x^2 - 7x - 44 = 0

    (ii) Hence state the solution of 2log⁡3(x+1)−log⁡3(x+5)=22\log_3(x + 1) - \log_3(x + 5) = 2, giving a reason why the other root of the quadratic must be rejected.

    (2 marks)
    Level 3 · 5 marksTry it

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