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

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Answer 1 · 2016-07-03T01:01:33

Use the log rule $log_a(m) + log_a(n) = log_a(m xx n)$

$ln(x(x + 2) = 1$

$ln(x^2 + 2x) = 1$

$x^2 + 2x = e^1$

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

$x = (-b +- sqrt(b^2 - 4ac))/(2a)$

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

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

$x = 0.93 and -2.93$

However, checking the solutions in the original equation we find that $x= -2.93$ is **extraneous. **

Hence, the solution set is ${0.93}$ .

Hopefully this helps!

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