A real root of the equation log_{4}{log_{2}(s | vedclass

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(A) Given equation: $\log_{4}\{\log_{2}(\sqrt{x+8} - \sqrt{x})\} = 0$Using the definition of logarithm,$\log_{b}(a) = c \Rightarrow a = b^{c}$:$\log_{

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(A) Given equation: $\log_{4}\{\log_{2}(\sqrt{x+8} - \sqrt{x})\} = 0$Using the definition of logarithm,$\log_{b}(a) = c \Rightarrow a = b^{c}$:$\log_{2}(\sqrt{x+8} - \sqrt{x}) = 4^{0} = 1$Again,using the definition of logarithm:$\sqrt{x+8} - \sqrt{x} = 2^{1} = 2$Rearranging the terms:$\sqrt{x+8} = 2 + \sqrt{x}$Squaring both sides:$x + 8 = (2 + \sqrt{x})^{2}$$x + 8 = 4 + x + 4\sqrt{x}$Subtracting $x$ and $4$ from both sides:$4 = 4\sqrt{x}$$1 = \sqrt{x}$Squaring both sides again:$x = 1$Verification: For $x=1$,$\sqrt{1+8} - \sqrt{1} = 3 - 1 = 2$. Then $\log_{2}(2) = 1$,and $\log_{4}(1) = 0$. The root is valid.

$A$ real root of the equation $\log_{4}\{\log_{2}(\sqrt{x+8} - \sqrt{x})\} = 0$ is

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