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1.8 Exponentials and Logarithms

1.8 Exponentials and Logarithms

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Question 2
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

[2]
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.

[4]
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1.8 Exponentials and Logarithms Questions

  1. A Level
  2. /Maths
  3. /1.8 Exponentials and Logarithms

104 exam-style questions on OCR (MEI) A Level Maths 1.8 Exponentials and Logarithms, covering 1.8.1 The function y = a^x, 1.8.2 Convert between index and logarithmic form, 1.8.3 Logarithm as inverse of exponential, 1.8.4 Laws of logarithms, 1.8.5 Values of log_a a and log_a 1, 1.8.6 Solve equations of the form a^x = b, 1.8.7 Reduce y = ax^n and y = ab^x to linear form, 1.8.8 The function y = e^x, 1.8.9 Gradient of e^kx (A-level only), 1.8.10 The function y = ln x, and 1.8.11 Exponential growth and decay. Each one has a worked solution and a mark scheme showing where the marks go.

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