How do you solve $ln x + ln (x+3) = 1$ ?

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Answer 1 · 2016-09-25T03:57:54

$x~=0.72896$

$lnx+ln(x+3)=1$

$ln(x*(x+3))=1$

$x(x+3) = e^1 = e$

$x^2+3x-e=0$

$x=(-3+-sqrt(9+4e))/2$

$~= (-3+-sqrt(9+10.87313))/2$

$~=(-3+-4.45793)/2$

Since $lnx$ is defined for $x>0$

$x~= (-3+4.45793)/2$

$x~=0.72896$

How do you solve ln x + ln (x+3) = 1ln x + ln (x+3) = 1?