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Question

If $$\log _{ 0.3 }{ (x-1) } <\log _{ 0.09 }{ (x-1) } $$, then find the interval in which $$x$$ lies.


A
x>2
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B
x<2
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C
x>2
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D
None of these
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Solution

The correct option is A $$x> 2$$
$$ log_{0.3}(x-1)<log_{0.09}(x-1) $$ 
$$\Rightarrow \log_{0.3}(x-1) < \dfrac{\log_{0.3}(x-1)}{\log_{0.3}(0.3)^2}$$
$$\Rightarrow \log_{0.3}(x-1) <\dfrac{1}{2} \log_{0.3}(x-1)$$
$$\Rightarrow \log_{0.3}(x-1) < \log_{0.3}\sqrt{x-1}$$
Taking antilog
$$\Rightarrow (x-1)>\sqrt{x-1}$$
$$\Rightarrow (x-1)^2-(x-2) >0$$
$$ \Rightarrow (x-1)(x-2)>0 $$ 
$$As \boxed{x\neq 1}$$
$$\Rightarrow \boxed{x>2}$$

1085276_516624_ans_b715fadbfd8843fba315436601ff0528.png

Mathematics

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