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

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

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$1$

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(A) Given the equation: $\log_{4}\{\log_{2}(\sqrt{x + 8} - \sqrt{x})\} = 0$By the definition of logarithm,$\log_{b}(a) = c \implies a = b^c$. Thus:$\log_{2}(\sqrt{x + 8} - \sqrt{x}) = 4^0 = 1$Again,applying 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$$x + 8 = 4 + x + 4\sqrt{x}$Subtracting $x$ from both sides:$8 = 4 + 4\sqrt{x}$$4 = 4\sqrt{x}$$1 = \sqrt{x}$Squaring both sides again:$x = 1$Checking the domain: For $\sqrt{x+8} - \sqrt{x}$ to be defined,$x \ge 0$. For $\log_{2}(\sqrt{x+8} - \sqrt{x})$ to be defined,$\sqrt{x+8} - \sqrt{x} > 0$,which is true for $x=1$ as $\sqrt{9} - 1 = 3 - 1 = 2 > 0$.

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

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