A real root of the equation {log _4}{ {log _2 | vedclass

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Question

(a) ${\log _4}\left\{ {\,{{\log }_2}(\sqrt {x + 8} - \sqrt x )} \right\} = 0$

$ \Rightarrow {4^0} = {\log _2}\left( {\sqrt {x + 8} - \sqrt x } \rig

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(a) ${\log _4}\left\{ {\,{{\log }_2}(\sqrt {x + 8} - \sqrt x )} \right\} = 0$

$ \Rightarrow {4^0} = {\log _2}\left( {\sqrt {x + 8} - \sqrt x } \right)$$ \Rightarrow {2^1} = \sqrt {x + 8} - \sqrt x $

$ \Rightarrow 4 = x + 8 + x - 2\sqrt {{x^2} + 8x} $

$ \Rightarrow 2\sqrt {{x^2} + 8x} = 2x + 4$ $ \Rightarrow {x^2} + 8x = {x^2} + 4 + 4x$

$ \Rightarrow 4x = 4$ $ \Rightarrow x = 1$.

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

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