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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?

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    $\begingroup$ $\ln a + \ln b = \ln (ab)$, not $\ln (a+b)$ or $\ln (a/b)$. Also you need $\pm$ when you remove the square root from both sides of the equation. $\endgroup$ – player3236 Jan 18 at 14:45
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    $\begingroup$ Start with $\ln(9 (x^2-10))$ on the LHS. $\endgroup$ – David Mitra Jan 18 at 14:45
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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$$\ln(x^2-10)+\ln(9)=\ln\left(\frac{x^2-10}9\right).$$

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    $\begingroup$ Note for OP: the step where we use the one-to-one property of logs is when we go from $\ln(9(x^2-10))=\ln(10)$ to $9x^2 - 90 = 10$. This step comes from performing the inverse function $\exp$ on both sides (doing $e$ to the power of both sides). That the function $\log$ has an inverse is a direct consequence of the fact that it is one-to-one. $\endgroup$ – Joe Jan 18 at 14:53
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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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