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If $\log _{0.3}(x-1) < \log _{0.09}(x-1),$ then $x$ lies in
the interval
Options:
the interval
Solution:
2494 Upvotes
Verified Answer
The correct answer is:
$(2, \infty)$
Given, $\quad \log _{0.3}(x-1) < \log _{0.09}(x-1)$
$\Rightarrow \quad \log _{0.3}(x-1) < \log _{0.3^{2}}(x-1)$
$\Rightarrow \quad \log _{0.3}(x-1)^{2} < \log _{0.3}(x-1)$
$\Rightarrow \quad(x-1)^{2}>x-1 \quad[\because 0.3 < 1]$
$\Rightarrow x^{2}+1-2 x-x+1>0$
$\Rightarrow \quad x^{2}-3 x+2>0$
$\Rightarrow \quad(x-1)(x-2)>0$
$\Rightarrow \quad x < 1, x>2 \Rightarrow x>2 \quad[\because x \neq 1]$
$\therefore \quad(2, \infty)$
Hence, x les in the interval $(2, \infty)$
$\Rightarrow \quad \log _{0.3}(x-1) < \log _{0.3^{2}}(x-1)$
$\Rightarrow \quad \log _{0.3}(x-1)^{2} < \log _{0.3}(x-1)$
$\Rightarrow \quad(x-1)^{2}>x-1 \quad[\because 0.3 < 1]$
$\Rightarrow x^{2}+1-2 x-x+1>0$
$\Rightarrow \quad x^{2}-3 x+2>0$
$\Rightarrow \quad(x-1)(x-2)>0$
$\Rightarrow \quad x < 1, x>2 \Rightarrow x>2 \quad[\because x \neq 1]$
$\therefore \quad(2, \infty)$
Hence, x les in the interval $(2, \infty)$
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