(A) Given, $\log _{0.3}(x-1) 0$, which implies $x > 1$.We can rewrite the in

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(A) Given, $\log _{0.3}(x-1) For the logarithm to be defined, we must have $x-1 > 0$, which implies $x > 1$.We can rewrite the inequality as $\log _{0.3}(x-1) Using the property $\log _{a^n} b = \frac{1}{n} \log_a b$, we get $\log _{0.3}(x-1) Multiplying by $2$, we get $2 \log _{0.3}(x-1) Subtracting $\log _{0.3}(x-1)$ from both sides, we get $\log _{0.3}(x-1) Since the base $0.3 (0.3)^0$.$x-1 > 1$.$x > 2$.Thus, $x$ lies in the interval $(2, \infty)$.

Class 11 Mathematics Basic of Logarithms Logarithms

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

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A
$(2, \infty)$
B
$(1, 2)$
C
$(-2, -1)$
D
None of these

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

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