Question

How do you solve `1/(log_2x-4)>1/(log_2x)`?

My solution:
`1/(log_2x-4)>1/(log_2x) => (log_2x-(log_2x-4))/((log_2x-4)log_2x) => (log_2x-log_2x+4)/((log_2x-4)log_2x) => (4)/((log_2x-4)log_2x)`
`log_2x!=4 => log_2x!=log_(2)2^4 => x!=16`
`log_2x!=0 => log_2x!=log_(2)2^0 => x!=1`
Answer:
`(-oo;1)uu(16;+oo)`

Answer 1

The solution is `x in (0,1)uu(16,+oo)`

Explanation:

The function `logx` is defined for

`x in (0,+oo)`

`f(x)=4/((log_2x)(log_2x-4))`

Make a sign chart

`color(white)(aaaaa)` `x` `color(white)(aaaaaa)` `0` `color(white)(aaaaaaa)` `1` `color(white)(aaaaaaa)` `16` `color(white)(aaaaa)` `+oo`

`color(white)(aaaa)` `log_2x` `color(white)(aaaaaaa)` `-` `color(white)(aa)` `||` `color(white)(aaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `log_2x-4` `color(white)(aa)` ###color(white)(aa)#`-` `color(white)(aaaaa)` `-` `color(white)(aa)` `||` `color(white)(aa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaaa)` `+` `color(white)(aa)` `||` `color(white)(aa)` `-` `color(white)(aa)` `||` `color(white)(aa)` `+`

Therefore,

`f(x)<0` when `x in (0,1)uu(16,+oo)`