Answer
$$x=1$$
Work Step by Step
$\ln{(x^2+1)}-3\ln{x}=\ln2$
$\ln{(x^2+1)}-\ln{x^3}=\ln2$
Writing the left side as the natural log of a single expression:
$\ln{(\frac{x^2+1}{x^3})}=\ln2$
Raising $e$ to the power of both sides:
$\frac{x^2+1}{x^3}=2$
Thus:
$x^2+1=2x^3$
$2x^3-x^2-1=0$
Factorizing the left side of the equation:
$(x-1)(2x^2+x+1)=0$
If we attempt to find the solution to $2x^2+x+1=0$, simply calculating the discriminant of the quadratic equation yields a negative number (-7), which demonstrates that there are no real solutions for $x$.
Therefore, $x=1$.
