If {log _{0.04}}(x - 1) ge {log _{0.2}}(x - 1 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the inequality: ${\log _{0.04}}(x - 1) \ge {\log _{0.2}}(x - 1)$.First,for the logarithm to be defined,we must have $x - 1 > 0$,which implie

Vedclass · Accepted Answer

$\left[ {2, + \infty } \right)$

Vedclass · Answer

(C) Given the inequality: ${\log _{0.04}}(x - 1) \ge {\log _{0.2}}(x - 1)$.First,for the logarithm to be defined,we must have $x - 1 > 0$,which implies $x > 1$.We can rewrite the base $0.04$ as $(0.2)^2$. Thus,${\log _{(0.2)^2}}(x - 1) \ge {\log _{0.2}}(x - 1)$.Using the property ${\log _{a^n}}b = \frac{1}{n}{\log _a}b$,we get $\frac{1}{2}{\log _{0.2}}(x - 1) \ge {\log _{0.2}}(x - 1)$.Rearranging the terms: $0 \ge {\log _{0.2}}(x - 1) - \frac{1}{2}{\log _{0.2}}(x - 1)$,which simplifies to $0 \ge \frac{1}{2}{\log _{0.2}}(x - 1)$.This implies ${\log _{0.2}}(x - 1) \le 0$.Since the base $0.2 $x - 1 \ge 1$,which gives $x \ge 2$.Combining this with the domain condition $x > 1$,we get $x \in [2, +\infty)$.

If ${\log _{0.04}}(x - 1) \ge {\log _{0.2}}(x - 1)$,then in which interval does $x$ lie?

Similar Questions

If $\log _{e}\left(x^{2}-16\right) \leq \log _{e}(4 x-11)$, then

If $\log_{4}5 = a$ and $\log_{5}6 = b$,then $\log_{3}2$ is equal to

The set of real values of $x$ satisfying $\log _{1/2}(x^2 - 6x + 12) \ge -2$ is

The real root of the equation ${\log _4}\{ {\log _2}(\sqrt {x + 8} - \sqrt x )\} = 0$ is..........

If $a^x = b^y = c^z = d^w$, the value of $x\left(\frac{1}{y} + \frac{1}{z} + \frac{1}{w}\right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module