Reverse Logarithmic Inequality

Relevant wiki: Logarithmic Inequalities - Problem Solving

In L.H.S the domain is $(-\infty,-2) \cup (-1,\infty)$ . In R.H.S the domain is $(-3,-2) \cup (-2,\infty)$ . So domain of $x$ is $(-3,-2) \cup (-1,\infty)$ .

!!We have $\log_{4}\left(\dfrac{x+1}{x+2}\right)>\log_{4}(x+3)$ or $\left(\dfrac{x+1}{x+2}\right)-(x+3)>0$ or $\dfrac{-(x^2+4x+5)}{x+2}>0$ .

$x^2+4x+5>0$ for all real $x$ as its discriminant is less than $0$ . So we need $x+2<0$ or $x<-2$ .!!

But the domain of $x$ is $(-3,-2) \cup (-1,\infty)$ !!so the domain of $x$ satisfying the inequality is also $(-1,\infty)$ . !!

Relevant wiki: Logarithmic Inequalities - Problem Solving

First we need to convert this question using a simple log formula :- $\log _{b}{a}=\frac{1}{\log _{a}{b}}$ Therefore, the above inequality becomes :

$\log _{ \frac { x+1 }{ x+2 } }{ 4 } < \log _{x+3}{4}$

Now, to find the domain of the LHS

$\frac { x+1 }{ x+2 } > 0$ on solving we get: $x\in \left( -\infty ,-2 \right) \cup\left( -1,\infty \right)$

Similiarly on solving for the domain of the RHS we get: $x \in (-3, \infty )$

Therefore the domain of x will be the intersection of both the sets i.e : $x \in (-3,-2) \cup(-1,\infty )$

However we need to find only those values of x for which this inequality is satisfied. We know that, $\log _{ y }{ 4 }$ is a monotonically decreasing function for y>1. For us to be able to use this monotonicity both $\frac { x+1 }{ x+2 }$ and $x+3$ must be greater than 1, which will not happen in the domain of $x$ .

Therefore , the only case is that $\frac {x+1}{x+2}< 1$ & $x+3 > 1$ . The only common solution satisfying both the conditions is $x > -2$ .The solution to this inequality will be the intersection of $\left(-2,\infty \right)$ and the domain of $x$ and hence the answer is $\boxed{(-1, \infty)}$

Sorry, if monotonicity was a bit confusing, I thought to use that concept to make this inequality a bit easier. Anyways do feel free to check out the graph of $\log_x 4$ at this link and figure this question out yourself!

LHS defined for $x \in (-\infty, -2) \cup (-1, \infty)$ and RHS defined for $(-3, -2) \cup (-2, \infty)$ . Hence the domain of $x$ is $(-3, -2) \cup (-1, \infty)$ .

To use the reciprocal law, both sides must be positive or negative. LHS is positive for $(-3, -2)$ while RHS is negative. For $(-1, \infty)$ , LHS is negative while RHS is positive. Hence, that's the solution.