Use the one to one property of logs to solve $\ln(x^2-10)+\ln(9)=\ln(10)$ Use the one to one property of logs to solve $\ln(x^2-10)+\ln(9)=\ln(10)$.
I get $x=\sqrt{11}$ or $x=10$ whereas my textbook says it's $x=\pm\frac{10}{3}$.
My working - initial attempt:
$$x^2-10+9=10$$
$$x^2-1=10$$
$$x^2=11$$
$$x=\sqrt{11}$$
My working - another attempt:
$$\ln(\frac{x^2-10}{9})=\ln(10)$$
$$\frac{x^2-10}{9}=10$$
$$x^2-10=90$$
$$x^2=100$$
$$x=10$$
Where am I going wrong? How can I arrive at $x=\pm\frac{10}{3}$? Don't both my solutions make sense? Why are they incorrect?

Note that\begin{align*}\ln(x^2-10)+\ln(9)=\ln(10)&\iff\ln\bigl(9(x^2-10)\bigr)=\ln(10)\\&\iff9x^2-90=10\\&\iff x^2=\frac{100}9\\&\iff x=\pm\frac{10}3.\end{align*}
The error from your first attempt lies in assuming that$$\ln(x^2-10)+\ln(9)=\ln(10)\iff x^2-10+9=10,$$whereas the error in your second attempt lies in assuming that$$\ln(x^2-10)+\ln(9)=\ln\left(\frac{x^2-10}9\right).$$
