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

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Answer 1 · 2015-09-10T16:28:21

Use exponentiation to produce a quadratic equation to solve using the quadratic formula.

Discard the root of this quadratic that results in #x < 0#, leaving the solution:

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

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

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

Then using the quadratic formula, with #a = 1#, #b = -2# and #c = e#

#x = (-b +-sqrt(b^2-4ac))/(2a) = (2 +-sqrt(2^2-(4xx1xx-e)))/(2*1)#

#=(2+-sqrt(4+4e))/2#

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

Now #sqrt(1+e) > 1#, so #1 - sqrt(1+e) < 0# and #ln(1 - sqrt(1+e))# is not defined.

So the only valid solution is #x = 1 + sqrt(1+e)#