We are seeking the solution of the inequality:
[tex]\displaystyle{ \log_{0.8}(x+4)\ \textgreater \ \log_{0.4}(x+4)[/tex].
We recall that a log function [tex]f(x)=\log_b(x)[/tex] is either increasing or decreasing:
i) it is increasing if b>1,
ii) it is decreasing if 0<b<1.
Consider the functions [tex]\displaystyle{ \log_{0.8}(x)[/tex] and [tex]\displaystyle{ \log_{0.4}(x)[/tex].
The graphs of these functions both meet at x=1 (clearly), and after 1 they are both negative. So from 0 to 1 one of them is larger for all x, and from 1 to infinity the other is larger. (Being strictly decreasing, their graphs can only intersect once.)
We can check for a certain convenient point, for example x=0.8:
[tex]\displaystyle{ \log_{0.8}(0.8)=1[/tex] and
[tex]\displaystyle{ \log_{0.4}(0.8)=\log_{0.4}(0.4\cdot 2)=\log_{0.4}(0.4)+\log_{0.4}(\cdot 2)=1+\log_{0.4}(2)[/tex].
Now, [tex]\displaystyle{ \log_{0.4}(2)[/tex] is negative since we already explained that for x>1 both functions were negative. This means that
[tex]\displaystyle{ \log_{0.8}(0.8)\ \textgreater \ \log_{0.4}(0.8)[/tex], and since [tex]0.8\in (0, 1)[/tex], then this is the interval where [tex]\displaystyle{ \log_{0.8}(x)\ \textgreater \ \log_{0.4}(x)[/tex].
So, now considering the functions [tex]\displaystyle{ \log_{0.8}(x+4)[/tex] and [tex]\displaystyle{ \log_{0.4}(x+4)[/tex], we see that
x+4 must be in the interval (0,1), so we solve:
0<x+4<1, which yields -4<x<-3 after we subtract by 4.
Answer: (-4, -3). Attached is the graph generated using Desmos.
