## Calculus (3rd Edition)

$$x=1$$
$\ln{(x^2+1)}-3\ln{x}=\ln2$ $\ln{(x^2+1)}-\ln{x^3}=\ln2$ 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, $x=1$.