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?
$\endgroup$ 22 Answers
$\begingroup$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$$\ln(x^2-10)+\ln(9)=\ln\left(\frac{x^2-10}9\right).$$
$\endgroup$ 1 $\begingroup$Your first attempt simply uses the strategy "let's ignore $\ln$": but it does not work.
The second attempt is better, but you made a mistake. Actually$$\ln (x^2-10) + \ln 9$$becomes$$\ln ((x^2-10) \cdot 9)$$then you have$$(x^2-10) \cdot 9=10$$ which is easily solved.
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