Question

If \(\log _{2} 6+\frac{1}{2 x}=\log _{2}\left(2^{\frac{1}{x}}+8\right)\), then the values of \(x\) are
(A) \(\frac{1}{4}, \frac{1}{3}\)
(B) \(\frac{1}{4}, \frac{1}{2}\)
(C) \(-\frac{1}{4}, \frac{1}{2}\)
(D) \(\frac{1}{3}, \frac{1}{-2}\)

Answer 1

Ans: (B)
Hint : \(\log _{2} 6+\log _{2} 2^{\frac{1}{2 x}}=\log _{2}\left(2^{\frac{1}{x}}+8\right) \Rightarrow 6 \cdot 2^{\frac{1}{2 x}}=2^{\frac{1}{x}}+8\), let \(2^{\frac{1}{2 x}}=\mathrm{a}\)
$$
\Rightarrow a^{2}-6 a+8=0 \Rightarrow a=4, a=2 \Rightarrow x=\frac{1}{4}, \frac{1}{2}
$$

Answer 2

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Answer 3

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