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Changing a base-25 logarithm to base 5 and solving a logarithmic equation in cos x on an interval

Changing a base-25 logarithm to base 5 and solving a logarithmic equation in cos x on an interval

8 marks
  1. Show that \(\log_{25}{(4\cos{2x}+3)}=\log_{5}{\sqrt{4\cos{2x}+3}}\).[3]
  2. Find every value of \(x\) in the interval \(0<x<\frac{\pi}{2}\) that satisfies \(\log_{5}{(2\cos{x})}=\log_{25}{(4\cos{2x}+3)}\), using part (a) or any other method.[5]

Solution:

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Answer:(a) \(\log_{25}{(4\cos{2x}+3)}=\log_{5}{\sqrt{4\cos{2x}+3}}\) (b) \(x=\dfrac{\pi}{3}\)

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