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

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Answer 1 · 2016-09-01T15:30:40

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

Start by applying the rule #log_a(n) + log_a(m) = log_a(n xx m)#.

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

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

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

#x^2 - 2x = e#

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

#x = (-(-2) +- sqrt(-2^2 - (4 xx 1 xx -e)))/(2 xx 1)#

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

#x = (2 +- sqrt(4(1 + e)))/2#

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

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

However, the #-# solution is extraneous, since it renders the original equation undefined. So, the only actual solution is #x = 1+sqrt(1 + e)#

Hopefully this helps!