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

1 Answer

Jul 10, 2016

#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

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