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

1 Answer
Jun 15, 2016

Answer:

#x = 1+sqrt(1+e)#

Explanation:

Assuming we are looking for Real solutions, we want to find Real values of #x# such that:

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

Applying the exponential fundtion to both sides, we have:

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

Subtracting #e# from both ends we get:

#0 = x^2-2x-e#

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

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

#= ((x-1)-sqrt(e+1))((x-1)+sqrt(e+1))#

#= (x-(1+sqrt(e+1)))(x-(1-sqrt(e+1)))#

So #x=1+-sqrt(e+1)#

Note that #sqrt(e+1) > 1#, so:

  • #x = 1-sqrt(1+e) < 0# making the arguments to #ln# in the original equation negative, so having no Real value.

  • #x = 1+sqrt(1+e) > 2# yielding Real values for the natural logarithms in the original equation.

So #x = 1+sqrt(1+e)# is the only valid Real soltuion of the original equation.

#color(white)()#
Footnote

Note that #x = 1-sqrt(1+e)# is not a solution of the original equation, even if we allow Complex values for #ln#.

In general:

#ln(z) = ln abs(z) + Arg(z)i#

So in the case of Real negative values of #z# we find:

#ln(z) = ln abs(z) + pii#

In particular:

#ln(1-sqrt(1+e))+ln(1-sqrt(1+e)-2)#

#=ln(sqrt(1+e)-1) + pii + ln(sqrt(1+e)+1) + pii#

#=ln((sqrt(1+e)-1)(sqrt(1+e)+1)) + 2pii#

#=ln(e) + 2pii#

#=1 + 2pii != 1#