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

1 Answer
Dec 27, 2016

Answer:

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

Explanation:

#ln a + ln b = ln ab  ; ln(exp(x)) = x#

#ln x(x + 2) = ln (exp 1)#

#x(x + 2) = e#

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

#Delta = 4 + 4e#

#x = (-2 ± 2 sqrt (1 + e))/2#

#x = -1 ± sqrt (1 + e)#

#x = -1 ± 1,9282846855324670226473240752689...#

But we can't extract LN(negative number)

#x > 0#