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

The equation $\log_e x + \log_e(1 + x) = 0$ can be written as

If $a, b, c \neq 0$ and belong to the set $\{1, 2, 3, \ldots, 9\}$,then $\log _{10}\left(\frac{a+10 b+10^2 c}{10^{-4} a+10^{-3} b+10^{-2} c}\right)$ is equal to

Given that $a, b \in \{0, 1, 2, \ldots, 9\}$ with $a+b \neq 0$ and that $\left(a+\frac{b}{10}\right)^x = \left(\frac{a}{10}+\frac{b}{100}\right)^y = 1000$. Then, $\frac{1}{x} - \frac{1}{y}$ is equal to

If $\log _{10} 7 = 0.8451$,then the position of the first significant figure of $7^{-20}$ is

The solution to the equation ${9^x} - {2^{x + 1/2}} = {2^{x + 3/2}} - {3^{2x - 1}}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module