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Basic of Logarithms
hard
${2^{{{\log }_{\sqrt 2 }}(x - 1)}} > x + 5$ નું સમાધાન કરે તેવી $x$ ની વાસ્તવિક કિમતોનો ગણ મેળવો.
A
$( - \infty ,\, - 1) \cup (4, + \infty )$
B
$(4, + \infty )$
C
$( - 1,\,4)$
D
એકપણ નહી.
Solution
(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., $x > 1$
$\therefore x > 4 \Rightarrow x \in (4,\,\infty )$.
Standard 11
Mathematics
