Why is the domain of the function #ln((2x)/(2+x))# x<-2, x>0?

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Answer 1 · 2018-03-22T16:18:05

The range is #y in(-oo, ln2)uu(ln2, +oo)#

The domain is given

#x in (-oo, -2) uu(0, +oo)#

To find the range, proceed as follows

Let #y=ln((2x)/(2+x))#

Therefore,

#e^y=(2x)/(2+x)#

#(2e^y+xe^y)=2x#

#x(2-e^y)=2e^y#

#x=(2e^y)/(2-e^y)#

So,

#x# will have values when the denominator is #!=0#

#2-e^y!=0#

#e^y!=2#

#y!=ln2#

The range is #y in(-oo, ln2)uu(ln2, +oo)#

graph{(y-ln((2x)/(2+x)))(y-ln2)=0 [-14.24, 14.24, -7.12, 7.12]}