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Paper 1 · June 2018 · Logarithms and Exponentials

Solve the equation $e^{2x} = 4e^{2-x}$ , giving your answer in exact form.

A

x = \tfrac{1}{2}(2 + \ln 4)

B

x = 2 + \ln 4

C

x = \tfrac{1}{3}\ln 4 - 2

D

x = \tfrac{1}{3}(2 + \ln 4)

Explanation

Dividing gives

e^{2x-(2-x)}=e^{3x-2}=4

, so

3x-2=\ln4

and

x=\tfrac13(2+\ln4)

.

Derived from ZIMSEC Mathematics Paper 1, June 2018, Q1

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