How do you solve # ln(x+1) - 1 = ln(x-1)#?

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Answer 1 · 2016-05-08T05:08:34

#x = (e+1)/(e-1)=coth (1/2)=2.164#, nearly.

Rearranging,

#ln (x+1)-ln(x-1)=1=ln e#.

#ln(x+1)/(x-1)=ln e#

So, #(x+1)/(x-1)=e#

#x=(e+1)/(e-1)#

#=(e^(1/2)(e^(1/2)+e^(-1/2)))/(e^(1/2)(e^(1/2)-e^(-1/2)))#

#=(e^(1/2)+e^(-1/2))/(e^(1/2)-e^(-1/2))#

#=coth (1/2)#

#(e+1)/(e-1)=2.164#, nearly.