we are given
[tex]log_0._8(x+4)>log_0._4(x+4)[/tex]
We can use base change formula
[tex]log_a(b)=\frac{ln(b)}{ln(a)}[/tex]
so, we get
[tex]\frac{ln(x+4)}{ln(0.8)} >\frac{ln(x+4}{ln(0.4)}[/tex]
we can move right side term to left side
[tex]\frac{ln(x+4)}{ln(0.8)} -\frac{ln(x+4)}{ln(0.4)}>0[/tex]
[tex]ln(x+4)(\frac{1}{ln(0.8)} -\frac{1}{ln(0.4)})>0[/tex]
we know that
[tex](\frac{1}{ln(0.8)} -\frac{1}{ln(0.4)})<0[/tex]
so, other term should also be less than 0
then only it can be positive or greater than 0
[tex]ln(x+4)<0[/tex]
Let's assume it is equal
[tex]ln(x+4)=0[/tex]
take exponent both sides
[tex]e^{ln(x+4)}=e^0[/tex]
[tex]x+4=1[/tex]
[tex]x=-3[/tex]
and ln(x+4) is also undefined at x+4=0
x=-4
now, we can draw a number line and locate this value
so, we get
[tex]-4<x<-3[/tex]...........Answer
