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]}