First slide

Trigonometric Identities

Question

Equation $\log_{4} (3 - x) + \log_{0 .25} (3 + x) = \log_{4} (1 - x) + \log_{0 .25} (2 x + 1)$ has

Moderate

A

only one prime solution
B

two real solutions
C

no real solution
D

none of these

Solution

$\begin{array}{l} \log_{4} (3 - x) + \log_{0 .25} (3 + x) = \log_{4} (1 - x) + \log_{0 .25} (2 x + 1) \\ \Rightarrow \log_{4} (3 - x) - \log_{4} (3 + x) = \log_{4} (1 - x) - \log_{4} (2 x + 1) \\ \Rightarrow \log_{4} (3 - x) + \log_{4} (2 x + 1) = \log_{4} (1 - x) + \log_{4} (3 + x) \\ \Rightarrow (3 - x) (2 x + 1) = (1 - x) (3 + x) \\ \Rightarrow 3 + 5 x - 2 x^{2} = 3 - 2 x - x^{2} \\ \Rightarrow x^{2} - 7 x = 0 \\ \Rightarrow x = 0, 7 \end{array}$
Only x = 0 is the solution and x=7 is to be rejected.

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