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

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Answer 1 · 2016-11-16T07:46:17

Combine the left into one term using the rule $ln(ab)=lna+lnb$

$lnx+ln(x+3)=1$
$ln[(x)(x+3)]=1$
$ln(x^2+3x)=1$

Rewrite using the definition of log:
$e^1=x^2+3x$
$e=x^2+3x$

Now solve as a quadratic; first set one side equal to zero:
$0=x^2+3x-e$

Use quadratic formula:
$x=(-b+-sqrt(b^2-4ac))/(2a)$

$x=(-(3)+-sqrt(3^2-4(1)(e)))/(2(1))$
$x=(-3+-sqrt(9-4e))/2$

$xapprox(-3+-sqrt(-1.87))/2$

Because we take the square root of a negative number, there are no solutions to the equation.

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