The real root of the equation {log _4}{ {log | vedclass

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(A) Given the equation: ${\log _4}\{ {\log _2}(\sqrt {x + 8}  - \sqrt x )\} = 0$Applying the definition of logarithm,we get: ${\log _2}(\sqrt {x + 8}

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$1$

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(A) Given the equation: ${\log _4}\{ {\log _2}(\sqrt {x + 8}  - \sqrt x )\} = 0$Applying the definition of logarithm,we get: ${\log _2}(\sqrt {x + 8}  - \sqrt x ) = {4^0} = 1$Again,applying the definition of logarithm: $\sqrt {x + 8} - \sqrt x = {2^1} = 2$Squaring both sides: $(\sqrt {x + 8} - \sqrt x)^2 = {2^2}$$x + 8 + x - 2\sqrt {x(x + 8)} = 4$$2x + 8 - 2\sqrt {x^2 + 8x} = 4$$2x + 4 = 2\sqrt {x^2 + 8x}$Dividing by $2$: $x + 2 = \sqrt {x^2 + 8x}$Squaring both sides again: $(x + 2)^2 = x^2 + 8x$$x^2 + 4x + 4 = x^2 + 8x$$4 = 4x$$x = 1$

The real root of the equation ${\log _4}\{ {\log _2}(\sqrt {x + 8} - \sqrt x )\} = 0$ is..........

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