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1.6 Exponentials and Logarithms

1.6 Exponentials and Logarithms

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Question 8
a.

Using the substitution y=2xy = 2^xy=2x, show that the equation 4x−2x+2−5=04^x - 2^{x+2} - 5 = 04x−2x+2−5=0 can be written as y2−4y−5=0y^2 - 4y - 5 = 0y2−4y−5=0

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b.

Hence show that the equation 4x−2x+2−5=04^x - 2^{x+2} - 5 = 04x−2x+2−5=0 has x=log⁡25x = \log_2 5x=log2​5 as its only solution.

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Markscheme

1.6 Exponentials and Logarithms Questions

  1. A Level
  2. /Maths
  3. /1.6 Exponentials and Logarithms

104 exam-style questions on OCR A Level Maths 1.6 Exponentials and Logarithms, covering 1.6.1 Properties of the exponential function, 1.6.2 Gradient of e^(kx) (A-level only), 1.6.3 Definition of the logarithm, 1.6.4 The natural logarithm function, 1.6.5 ln x as inverse of e^x, 1.6.6 Laws of logarithms, 1.6.7 Equations involving exponentials, 1.6.8 Reduction to linear form, and 1.6.9 Modelling using exponential functions. Each one has a worked solution and a mark scheme showing where the marks go.

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