How do you find the domain and range of $y=log(2x-3)/(x-5)$ ?

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Answer 1 · 2018-06-24T08:42:56

The domain is $x in (3/2,5)uu(5,+oo)$ . The range is $y in RR$

The function is

$y=ln(2x-3)/(x-5)$

For the domain, there are $2$ points to consider

${(2x-3>0),(x-5!=0):}$

$=>$ , ${(x>3/2),(x!=5):}$

Therefore,

The domain is $x in (3/2,5)uu(5,+oo)$

For the range, calculate the following limits

$lim_(x->3/2)y=+oo$

$lim_(x->5^-)y=-oo$

$lim_(x->5^+)y=+oo$

$lim_(x->oo)y=0^+$

The range is $y in RR$

graph{ln(2x-3)/(x-5) [-5.73, 16.77, -4.24, 7.01]}

How do you find the domain and range of y=log(2x-3)/(x-5) y=log(2x-3)/(x-5) ?