WBJEE · Maths · Basic of Mathematics
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 & Solution
Correct Answer
(B) \(\frac{1}{4},\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
We have, \(\log _{2}^{6}+\frac{1}{2 x}=\log _{2}\left(2^{\frac{1}{x}}+8\right)\) \(\Rightarrow \quad 1+\log _{2}^{3}+\frac{1}{2 x}=\log _{2}\left(2^{\frac{1}{x}}+8\right)\) \(\Rightarrow \quad \frac{2^{\frac{1}{x}}+8}{3}=2^{1+\frac{1}{2 x}}\)…
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