How do you solve #lnx+ln(x+1)=2 #?
1 Answer
Mar 19, 2018
# x = 1/2+1/2sqrt(1+4e^2) #
Explanation:
We have:
# lnx+ln(x-1) = 2#
# :. ln(x(x-1)) = 2#
# :. x(x-1) = e^2#
# :. x^2-x- e^2 = 0#
# :. (x-1/2)^2-(1/2)^2- e^2 = 0#
# :. (x-1/2)^2 = 1/4+e^2 = 0#
# :. x-1/2 = +-sqrt(1/4(1+4e^2)) #
# :. x-1/2 = +-1/2sqrt(1+4e^2) #
# :. x = 1/2+-1/2sqrt(1+4e^2) #
Strictly speaking we require that both:
# x gt 0# and#x+1 gt 0#
So that the logarithms, in the original equation exist. Consequently we can negate one solution and we have:
# x = 1/2+1/2sqrt(1+4e^2) #
