${2^{{{\log }_{\sqrt 2 }}(x - 1)}} > x + 5$ નું સમાધાન કરે તેવી $x$ ની વાસ્તવિક કિમતોનો ગણ મેળવો.
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b
(b) \({2^{{{\log }_{\sqrt 2 }}(x - 1)}} > x + 5\)\( \Rightarrow \)\({(\sqrt 2 )^{2{{\log }_2}(x - 1)}} > x + 5\)
(b) \({2^{{{\log }_{\sqrt 2 }}(x - 1)}} > x + 5\)\( \Rightarrow \)\({(\sqrt 2 )^{2{{\log }_2}(x - 1)}} > x + 5\)
\( \Rightarrow \) \({(x - 1)^2} > x + 5\)\( \Rightarrow \)\({x^2} - 3x - 4 > 0\)
\( \Rightarrow \) \((x - 4)\,(x + 1) > 0\)\( \Rightarrow \)\(x > 4\) or \(x < - 1\)
But for \({\log _{\sqrt 2 }}(x - 1)\) to be defined, \(x - 1 > 0\) i.e., \(x > 1\)
\(\therefore x > 4 \Rightarrow x \in (4,\,\infty )\).
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