If log _{e}left(x^{2}-16right) leq log _{e}(4 | vedclass

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

Question

(A) For the logarithmic inequality $\log _{e}\left(x^{2}-16\right) \leq \log _{e}(4 x-11)$ to be defined,the arguments must be positive: $1) \ x^{2}-1

Vedclass · Accepted Answer

$4 < x \leq 5$

Vedclass · Answer

(A) For the logarithmic inequality $\log _{e}\left(x^{2}-16\right) \leq \log _{e}(4 x-11)$ to be defined,the arguments must be positive: $1) \ x^{2}-16 > 0$ $\Rightarrow (x-4)(x+4) > 0$ $\Rightarrow x \in (-\infty, -4) \cup (4, \infty)$ $2) \ 4x-11 > 0 \Rightarrow x > \frac{11}{4} = 2.75$ Combining these,the domain is $x > 4$. Now,solving the inequality: $x^{2}-16 \leq 4x-11$ $x^{2}-4x-5 \leq 0$ $(x-5)(x+1) \leq 0$ This holds for $x \in [-1, 5]$. Taking the intersection of the domain $(x > 4)$ and the solution $(x \in [-1, 5])$,we get $4 < x \leq 5$.

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

Similar Questions

The equation $x^{\frac{3}{4}(\log_2 x)^2 + \log_2 x - \frac{5}{4}} = \sqrt{2}$ has

For $y = \log_a x$ to be defined,$a$ must be:

The sum of the roots of the equation $x+1-2 \log _{2}\left(3+2^{x}\right)+2 \log _{4}\left(10-2^{-x}\right)=0$ is:

The number of solutions of the equation $\frac{1}{2} \log _{\sqrt{3}}\left(\frac{x+1}{x+5}\right)+\log _{9}(x+5)^{2}=1$ is

The equation $x^{(\log _{3} x)^{2}-\frac{9}{2} \log _{3} x+5}=3 \sqrt{3}$ has

Vedclass Test Series

Exam Paper Generator

Online Exam Module