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

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Answer 1 · 2017-01-16T14:45:48

$x=(sqrt(1+4e)-1)/2=1.223$

$lnx+ln(x+1)=1$

Hence $x(x+1)=e$

i.e. $x^2+x-e=0$

and using quadratic formula

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

$=(-1+-sqrt(1+4e))/2$

$=(-1+-3.446)/2$

$=(2.446)/2$ or $-4.446/2$

$=1.223$ or $-2.223$

But as $x=-2.223$ is not in domain, we have $x=(sqrt(1+4e)-1)/2=1.223$
graph{lnx+ln(x+1)-1 [-5, 5, -2.5, 2.5]}

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