text{If } log _{0.2}(x-1) log _{0.04}(x+5), | vedclass

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(C) We have, $\log _{0.2}(x-1) > \log _{0.04}(x+5)$ $\Rightarrow \log _{0.2}(x-1) > \log _{0.2^{2}}(x+5)$ $\Rightarrow \log _{0.2}(x-1) > \frac{1}{2}

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$1 < x < 4$

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(C) We have, $\log _{0.2}(x-1) > \log _{0.04}(x+5)$ $\Rightarrow \log _{0.2}(x-1) > \log _{0.2^{2}}(x+5)$ $\Rightarrow \log _{0.2}(x-1) > \frac{1}{2} \log _{0.2}(x+5)$ $\Rightarrow 2 \log _{0.2}(x-1) > \log _{0.2}(x+5)$ $\Rightarrow \log _{0.2}(x-1)^{2} > \log _{0.2}(x+5)$ $\Rightarrow (x-1)^{2} $[\because \log _{a} x > \log _{a} y \Rightarrow x $\Rightarrow x^{2}-2x+1 $\Rightarrow x^{2}-3x-4 $\Rightarrow (x-4)(x+1) $\Rightarrow x \in (-1, 4)$ Also, for the logarithm to be defined, $x-1 > 0 \Rightarrow x > 1$ and $x+5 > 0 \Rightarrow x > -5$. Combining $x \in (-1, 4)$ and $x > 1$, we get $x \in (1, 4)$.

$\text{If } \log _{0.2}(x-1) > \log _{0.04}(x+5), \text{ then }$

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