Question

For what values of x is log base 0.8 (x+4)>log base 0.4 (x+4)Thank you!

Answer 1

We are seeking the solution of the inequality:

$\displaystyle{ \log_{0.8}(x+4)\ \textgreater \ \log_{0.4}(x+4)$.


We recall that a log function $f(x)=\log_b(x)$ is either increasing or decreasing:

i) it is increasing if b>1, 

ii) it is decreasing if 0<b<1.

Consider the functions $\displaystyle{ \log_{0.8}(x)$ and $\displaystyle{ \log_{0.4}(x)$.

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:

$\displaystyle{ \log_{0.8}(0.8)=1$ and

$\displaystyle{ \log_{0.4}(0.8)=\log_{0.4}(0.4\cdot 2)=\log_{0.4}(0.4)+\log_{0.4}(\cdot 2)=1+\log_{0.4}(2)$.

Now, $\displaystyle{ \log_{0.4}(2)$ is negative since we already explained that for x>1 both functions were negative. This means that 

$\displaystyle{ \log_{0.8}(0.8)\ \textgreater \ \log_{0.4}(0.8)$, and since $0.8\in (0, 1)$, then this is the interval where $\displaystyle{ \log_{0.8}(x)\ \textgreater \ \log_{0.4}(x)$.


So, now considering the functions $\displaystyle{ \log_{0.8}(x+4)$ and $\displaystyle{ \log_{0.4}(x+4)$, 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.