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$\text {If } \log _{0.2}(x-1)>\log _{0.04}(x+5), \text { then }$
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Verified Answer
The correct answer is:
$1 < x < 4$
We have, $\log _{0.2}(x-1)>\log _{0.04}(x+5)$
$\Rightarrow \quad \log _{0.2}(x-1)>\log _{0.2^{2}}(x+5)$
$\Rightarrow \quad \log _{0.2}(x-1)>\frac{1}{2} \log _{0.2}(x+5)$
$\Rightarrow \quad 2 \log _{0.2}(x-1)>\log _{0.2}(x+5)$
$\Rightarrow \quad \log _{0.2}(x-1)^{2}>\log _{0.2}(x+5)$
$\Rightarrow \quad(x-1)^{2} < x+5$
$\left[\because \log _{a} x>\log _{a} y \Rightarrow x < y,\right.$ if $\left.0 < a < 1\right]$
$\Rightarrow \quad x^{2}-2 x+1 < x+5$
$\Rightarrow \quad x^{2}-3 x-4 < 0$
$\Rightarrow \quad x^{2}-4 x+x-4 < 0$
$\Rightarrow x(x-4)+1(x-4) < 0$
$\Rightarrow \quad(x-4)(x+1) < 0$
$\Rightarrow \quad x \in(-1,4)$
But $\quad x>1$
$\Rightarrow \quad x \in(1,4)$
$\Rightarrow \quad \log _{0.2}(x-1)>\log _{0.2^{2}}(x+5)$
$\Rightarrow \quad \log _{0.2}(x-1)>\frac{1}{2} \log _{0.2}(x+5)$
$\Rightarrow \quad 2 \log _{0.2}(x-1)>\log _{0.2}(x+5)$
$\Rightarrow \quad \log _{0.2}(x-1)^{2}>\log _{0.2}(x+5)$
$\Rightarrow \quad(x-1)^{2} < x+5$
$\left[\because \log _{a} x>\log _{a} y \Rightarrow x < y,\right.$ if $\left.0 < a < 1\right]$
$\Rightarrow \quad x^{2}-2 x+1 < x+5$
$\Rightarrow \quad x^{2}-3 x-4 < 0$
$\Rightarrow \quad x^{2}-4 x+x-4 < 0$
$\Rightarrow x(x-4)+1(x-4) < 0$
$\Rightarrow \quad(x-4)(x+1) < 0$
$\Rightarrow \quad x \in(-1,4)$
But $\quad x>1$
$\Rightarrow \quad x \in(1,4)$
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