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The equation _{4}(3-x)+_{0.25}(3+x)= _{4}(1-x)+_{0.25}(2x+1)

Question

The equation $lo g_{4} (3 - x) + lo g_{0.25} (3 + x) = lo g_{4} (1 - x) + lo g_{0.25} (2 x + 1)$ has

only one prime solution.

two real solutions.

no real solution(s).

none of these

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The equation

lo g_{4} (3 - x) + lo g_{0.25} (3 + x) = lo g_{4} (1 - x) + lo g_{0.25} (2 x + 1)

has

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Text solutionVerified

lo g_{4} (3 - x) + lo g_{0.25} (3 + x) = lo g_{4} (1 - x) + lo g_{0.25} (2 x + 1)

Above equation is valid when,

3 - x > 0, 3 + x > 0, 1 - x > 0, 2 x + 1 > 0

\Rightarrow x \in (- \frac{1}{2}, 1)

lo g_{4} (3 - x) + lo g_{0.25} (3 + x) = lo g_{4} (1 - x) + lo g_{0.25} (2 x + 1)

\Rightarrow lo g_{4} (3 - x) - lo g_{4} (3 + x) = lo g_{4} (1 - x) - lo g_{4} (2 x + 1) [∵ lo g_{a^{n}} b^{m} = \frac{m}{n} lo g_{a} b]

\Rightarrow lo g_{4} (\frac{3 - x}{3 + x}) = lo g_{4} (\frac{1 - x}{2 x + 1}) [∵ lo g a - lo g b = lo g \frac{a}{b}]

\Rightarrow (\frac{3 - x}{3 + x}) = (\frac{1 - x}{2 x + 1})

\Rightarrow x^{2} - 7 x = 0

\Rightarrow x = 0, 7

Since,

x \in (- \frac{1}{2}, 1)

Therefore,

x = 0

Ans: D

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Question Text	The equation $lo g_{4} (3 - x) + lo g_{0.25} (3 + x) = lo g_{4} (1 - x) + lo g_{0.25} (2 x + 1)$ has
Answer Type	Text solution:1
Upvotes	150