How do you solve #ln(x+1) - lnx = 2#?

Question

2460 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-06T22:22:32

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

#ln(x+1)-lnx = 2#

#ln((x+1)/x) = ln(e^2)#

#cancel(ln)((x+1)/x) = cancel(ln)(e^2)#

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

#x+1 = xe^2#

#1 = xe^2 - x#

common factor

#1 = x(e^2 - 1)#

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