Solve for $x$ : $\log_e(x^2-16)\lt \log_e(4x-11)$ $\log_e(x^2-16)\lt \log_e(4x-11)$
My attempt:
Since the base is $\gt 1$, we have from the above ,
$$x^2-16-4x+11\lt 0\\ \implies x^2-4x-5\lt 0\\ \implies(x-5)\cdot (x+1)\lt 0$$
If I say $(x-5)\gt 0 ;x+1\lt 0$ then the solution is $x\in (5, \infty)$
If I take  $(x-5)\lt 0 ;x+1\gt 0$  then the solution is $x\in (0,1)\bigcup (1,5)$
So , finally the solution is $$x\in (0,1)\bigcup (1,\infty ).$$
Is it correct or did I make any mistake ? I doubt it because this is not the given answer . The given answer is somehow $x\in (4,5).$
 A: There is no $x$ such that $x-5\gt 0$ and $x+1\lt 0$.
In the second case, you have to have $x-5\lt 0$ and $x+1\gt 0$, i.e. $x\lt 5$ and $x\gt \color{red}{-1}$.
Don't forget to have $x^2-16\gt 0$ and $4x-11\gt 0$.
A: from the definition of logarithm we get
$$(x-4)(x+4)>0$$ and $$x>\frac{11}{4}$$ thus we have $$x>4$$ and by the law of lagarithm we get
$$\ln(1)<\ln\left(\frac{4x-11}{x^2-16}\right)$$ thus we have to solve
$$x^2-4x-5<0$$
from here we get $$-1<x<5$$ and $$x>4$$ from above and finally
$$4<x<5$$
A: First, the trinomial $(x-5)(x+1)$ is negative for $x$ between its roots, i.e. for $x \in (-1, 5)$.
Next, you also want the logarithms to exist, i.e. for their arguments to be positive. This means that $x^2 - 16 >0$ and $4x - 11 > 0$. The first inequation gives you $x \in (-\infty, -4) \cup (4, \infty)$. The second gives $x \in (\frac {11} 4, \infty)$.
Intersecting all these intervals gives you the solution: $x \in (4,5)$.
A: Beware: $\log_e z$ is defined only for $z>0$.
The inequality is equivalent to the system of inequalities
$$
\begin{cases}
x^2-16>0 \\[6px]
4x-11>0 \\[6px]
x^2-16<4x-11
\end{cases}
$$
Since the last one becomes $(x-5)(x+1)<0$, you can write the solutions of the inequalities above as
$$
\begin{cases}
x<-4 \text{ or } x>4 \\[6px]
x>11/4 \\[6px]
-1<x<5
\end{cases}
$$
and the intersection of the three solutions sets is
$$
4<x<5
$$
