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1 Unit 1: Pure Mathematics A

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Question 61
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 Unit 1: Pure Mathematics A Questions

  1. A Level
  2. /Maths
  3. /1 Unit 1: Pure Mathematics A

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