If log _{2} 6 + frac{1}{2x} = log _{2} (2^{fr | vedclass

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(B) Given equation: $\log _{2} 6 + \frac{1}{2x} = \log _{2} (2^{\frac{1}{x}} + 8)$Using $\log _{2} 6 = \log _{2} (2 	imes 3) = 1 + \log _{2} 3$,we ge

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$\frac{1}{4}, \frac{1}{2}$

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(B) Given equation: $\log _{2} 6 + \frac{1}{2x} = \log _{2} (2^{\frac{1}{x}} + 8)$Using $\log _{2} 6 = \log _{2} (2 	imes 3) = 1 + \log _{2} 3$,we get:$1 + \log _{2} 3 + \frac{1}{2x} = \log _{2} (2^{\frac{1}{x}} + 8)$$\log _{2} 3 + \frac{1}{2x} = \log _{2} (2^{\frac{1}{x}} + 8) - 1$$\log _{2} 3 + \frac{1}{2x} = \log _{2} (2^{\frac{1}{x}} + 8) - \log _{2} 2$$\log _{2} 3 + \frac{1}{2x} = \log _{2} \left( \frac{2^{\frac{1}{x}} + 8}{2} ight)$$\frac{1}{2x} = \log _{2} \left( \frac{2^{\frac{1}{x}} + 8}{2 	imes 3} ight) = \log _{2} \left( \frac{2^{\frac{1}{x}} + 8}{6} ight)$$2^{\frac{1}{2x}} = \frac{2^{\frac{1}{x}} + 8}{6}$Let $y = 2^{\frac{1}{2x}}$,then $y^2 = 2^{\frac{1}{x}}$.$6y = y^2 + 8$$y^2 - 6y + 8 = 0$$(y - 4)(y - 2) = 0$$y = 4$ or $y = 2$If $2^{\frac{1}{2x}} = 4 = 2^2$,then $\frac{1}{2x} = 2 \Rightarrow x = \frac{1}{4}$.If $2^{\frac{1}{2x}} = 2 = 2^1$,then $\frac{1}{2x} = 1 \Rightarrow x = \frac{1}{2}$.

If $\log _{2} 6 + \frac{1}{2x} = \log _{2} (2^{\frac{1}{x}} + 8)$,then the values of $x$ are

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