The set of real values of x for which {2^{{{l | vedclass

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Question

(b) ${2^{{{\log }_{\sqrt 2 }}(x - 1)}} > x + 5$$ \Rightarrow $${(\sqrt 2 )^{2{{\log }_2}(x - 1)}} > x + 5$

$ \Rightarrow $ ${(x - 1)^2} >

Vedclass · Accepted Answer

$(4, + \infty )$

Vedclass · Answer

(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., $x > 1$

$	herefore x > 4 \Rightarrow x \in (4,\,\infty )$.

The set of real values of $x$ for which ${2^{{{\log }_{\sqrt 2 }}(x - 1)}} > x + 5$ is

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