Question

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

Answer 1

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

Explanation:

`log_e(x+1) - 1 = log_e(x-1)->log_e((x+1)/e)=loge_e(x-1)->(x+1)/e=(x-1)`

then

`x =(e+1)/(e-1) > 1->`feasible solution