The correct answer is **D**. Let's solve the given equation: $$\frac{1}{\log_x 2} + \frac{1}{\log_{2x} 2} + \frac{1}{\log_{4x} 2} = 1$$ We begin by applying the change of base formula for logarithms, which states that $\log_b a = \frac{1}{\log_a b}$. This property allows us to convert the terms with variable bases to a common base, typically 2 in this case, as 2 appears frequently. Applying this property to each term: 1. The first term: $$\frac{1}{\log_x 2} = \log_2 x$$ 2. The second term: $$\frac{1}{\log_{2x} 2} = \log_2 (2x)$$ Now, we use the product rule of logarithms, $\log_b (MN) = \log_b M + \log_b N$: $$\log_2 (2x) = \log_2 2 + \log_2 x$$ Since $\log_2 2 = 1$, this simplifies to: $$\log_2 (2x) = 1 + \log_2 x$$ 3. The third term: $$\frac{1}{\log_{4x} 2} = \log_2 (4x)$$ Again, using the product rule of logarithms: $$\log_2 (4x) = \log_2 4 + \log_2 x$$ Since $\log_2 4 = \log_2 (2^2) = 2$, this simplifies to: $$\log_2 (4x) = 2 + \log_2 x$$ Now, substitute these simplified expressions back into the original equation: $$\log_2 x + (1 + \log_2 x) + (2 + \log_2 x) = 1$$ To simplify the algebra, let $y = \log_2 x$. The equation becomes: $$y + (1 + y) + (2 + y) = 1$$ Combine the like terms: $$(y + y + y) + (1 + 2) = 1$$ $$3y + 3 = 1$$ Subtract 3 from both sides of the equation: $$3y = 1 - 3$$ $$3y = -2$$ Divide by 3 to solve for $y$: $$y = -\frac{2}{3}$$ Finally, substitute back $y = \log_2 x$: $$\log_2 x = -\frac{2}{3}$$ To find the value of $x$, convert this logarithmic equation into its equivalent exponential form. The definition of a logarithm states that if $\log_b a = c$, then $a = b^c$. Applying this: $$x = 2^{-2/3}$$ This value of $x$ satisfies the given conditions ($x > 0$ and $x \neq 1, x \neq \frac{1}{2}, x \neq \frac{1}{4}$). Comparing our result with the given options: A. $2^{1/3}$ B. $2^{2/3}$ C. $2^{-1/3}$ D. $2^{-2/3}$ The calculated value of $x$ matches option D.