Share 45 in the ratio 1:2.
1 answer:
You might be interested in
We are seeking the solution of the inequality:
.
We recall that a log function
is either increasing or decreasing:
i) it is increasing if b>1,
ii) it is decreasing if 0<b<1.
Consider the functions
and 
.
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:
and
.
Now,
is negative since we already explained that for x>1 both functions were negative. This means that
, and since 
, then this is the interval where 
.
So, now considering the functions
and 
, 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.
We recall that a log function
i) it is increasing if b>1,
ii) it is decreasing if 0<b<1.
Consider the functions
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:
Now,
So, now considering the functions
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.