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) #