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The equation $\ln(|y+1|)= x-2$ where you solve for $y$, I am just unsure of how the absolute value plays into this. I am assuming that I would convert to exponential form to get $|y+1|=e^{x-2}$ and then I am stuck. Sorry if this is an easy question and I'm just missing something obvious. :/

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  • $\begingroup$ You're in the right track. Now that you've got to $|y+1| = e^{x-2}$, consider the two possible values of $|y+1|$ and solve for $y$ in each case. You'll end up with two possible solutions in the end. Hint: the definition of $|a|$: $|a| = a$ when $a \ge 0$ and $|a| = -a$ when $a < 0$, for any real number $a$. $\endgroup$ – wltrup Aug 4 '15 at 22:40
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Notice, $$\ln|y+1|=x-2$$ $$|y+1|=e^{x-2}$$ Since, $e^{x-2}>0 \ \forall\ x\in R$ $$ y+1=\pm e^{x-2}$$

$$\bbox[5px, border:2px solid #C0A000]{y=\pm e^{x-2}-1}$$

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  • $\begingroup$ $y = -1 - e^{x-2}$ is also a solution. $\endgroup$ – wltrup Aug 4 '15 at 22:43
  • $\begingroup$ Hah... better now. I'll take away my down vote now. $\endgroup$ – wltrup Aug 4 '15 at 22:46
  • $\begingroup$ yes,, you are absolutely right $\endgroup$ – Harish Chandra Rajpoot Aug 4 '15 at 22:46
  • $\begingroup$ so the whole inequality thing is not needed? $\endgroup$ – Legnd22 Aug 4 '15 at 23:18

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