What are possible values of x if $ln(x-4) +ln(3) <=0$ ?

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Answer 1 · 2018-05-14T04:39:58

Possible values of $x$ are given by $4< x<=13/3$

We can write $ln(x-4)+ln3<=0$ as

$ln(3(x-4))<=0$

graph{lnx [-10, 10, -5, 5]}

Now as $lnx$ is a function which always increases as $x$ increases (graph shown above) as also that $ln1=0$ , this means

$3(x-4)<=1$

i.e. $3x<=13$

and $x<=13/3$

Observe that as we have $ln(x-4)$ domain of $x$ is $x>4$

Hence possible values of $x$ are given by $4< x<=13/3$

What are possible values of x if ln(x-4) +ln(3) <=0ln(x-4) +ln(3) <=0?