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Given this equation: $$\ln(x^2-1)-3=\ln(x+1)$$

Evaluate x.

Applying the natural logarithmic rule

$$\ln(x+1)=3$$

$$x+1=e^3$$

$$x=e^3-1$$

The answer was different from the book. Where did I when wrong?

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    $\begingroup$ How did you get $\ln(x+1) = 3$??? There's no reason to think $ln(x^2 -1) -3 = 3$ so why does $\ln(x+1)=3$. $\endgroup$
    – fleablood
    Commented Jun 9, 2019 at 4:27

2 Answers 2

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You wrote $\ln(x+1)=3$ instead of $\ln(x-1)=3$.

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$\ln(x^2-1)=\ln((x+1)(x-1))=\ln(x+1)+\ln(x-1)=\ln(x+1)+3$

implies $\ln(x-1)=3$, $x=e^3+1$

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