Name: Logarithmic Inequalities
Uploaded: 2019-07-19
Duration: 10 min 24 s
Description: Logarithmic inequalities are inequalities in which one (or both) sides involve a logarithm. Watch this video and learn how to solve logarithmic inequalities.

Question

Find domain of the following functions $f (x) = {(x + 0.5)}^{^{l o g_{(0.5 + x)} \frac{x^{2} + 2 x - 3}{4 x^{2} - 4 x - 3}}}$ .

Accepted Answer

Given, $f (x) = {(x + 0.5)}^{^{l o g_{(0.5 + x)} \frac{x^{2} + 2 x - 3}{4 x^{2} - 4 x - 3}}}$

$\log$ is defined only when,

$\frac{x^{2} + 2 x - 3}{4 x^{2} - 4 x - 3} > 0 & 0.5 + x > 0 & 0.5 + x \neq 1$

$\Rightarrow \frac{x^{2} + 3 x - x - 3}{4 x^{2} + 2 x - 6 x - 3} > 0 & x > - \frac{1}{2} & x \neq \frac{1}{2}$

$\Rightarrow \frac{(x - 1) (x + 3)}{(2 x - 3) (2 x + 1)} > 0 & x > - \frac{1}{2} & x \neq \frac{1}{2}$

For, $\frac{(x - 1) (x + 3)}{(2 x - 3) (2 x + 1)} > 0$ ,

$\Rightarrow x \in (- \infty, - 3) \cup (- \frac{1}{2}, 1) \cup (\frac{3}{2}, \infty)$

Taking intersection of all above cases we get,

$x \in (- \frac{1}{2}, \frac{1}{2}) \cup (\frac{1}{2}, 1) \cup (\frac{3}{2}, \infty)$

We know that exponential functions are defined for all real values $x \in (- \infty, \infty)$ .

Hence, the domain of $f (x)$ is $x \in (- \frac{1}{2}, \frac{1}{2}) \cup (\frac{1}{2}, 1) \cup (\frac{3}{2}, \infty)$ .

Embibe Experts Solutions for Chapter: Logarithm and its Applications, Exercise 4: Exercise-4

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Logarithm and its Applications, Exercise 4: Exercise-4

Questions from Embibe Experts Solutions for Chapter: Logarithm and its Applications, Exercise 4: Exercise-4 with Hints & Solutions

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