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1.6 Exponentials and logarithms

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Question 6
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]
Markscheme

1.6 Exponentials and logarithms Questions

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

104 exam-style questions on WJEC A Level Maths 1.6 Exponentials and logarithms, covering 1.6.1 Exponentials and logarithms, 1.6.2 Exponentials and logarithms, 1.6.3 Exponentials and logarithms, 1.6.4 Exponentials and logarithms, 1.6.5 Exponentials and logarithms, 1.6.6 Exponentials and logarithms, 1.6.7 Exponentials and logarithms, 1.6.8 Exponentials and logarithms, and 1.6.9 Exponentials and logarithms. Each one has a worked solution and a mark scheme showing where the marks go.

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