Answer
$x=-1 + \sqrt {1+e^4} \approx 6.456$
Work Step by Step
Apply the logarithmic property:
$\log_a M+\log_a N = \log_a (MN)$
and rearrange the given expression to obtain:
$\ln[(x)(x+2)] \ ...(1)$
Since, $\log_m{n} = 1 $ gives: $m^{(1)}=n$; then we have:
$e^4=(x)(x+2)$
$x^2+2x=e^4$
or, $x^2+2x-e^4=0$
This is a quadratic equation; thus by using the quadratic formula $x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$, it will become:
$x=\dfrac{-(2) \pm \sqrt{(2)^2-(4)(1)(-e^4)}}{2(1)}=$
or, $x=-1 + \sqrt {1+e^4}, -1 - \sqrt {1+e^4}$
Since the domain of the variable is $x \gt 0$, we cannot accept the value of $x=-1 - \sqrt {1+e^4}$
Thus, our answer is: $x=-1 + \sqrt {1+e^4} \approx 6.456$
