Logarithmic Inequality with incomplete first attempt

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Thank you, that's a really interesting problem. I don't think I saw a similar one in any of my problem sets.
The way I understand it is that the initial inequality and the one with the 2 pulled out in front of the logarithm are not equivalent, because of different domains. If we look closer at the rule which tells us how to extract the exponent from the argument of a logarithm, it says the base has to be positive. But we can argue, that since (x+2)^2 = |x+2|^2, then in the most general case the base of the exponent in the logarithm was really |x+2| with the "hidden" absolute value. So the initial inequality is really equivalent to the 2*ln|x+2|<6 inequality

MrSzybciutki
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I feel like this video will help a lot of students. Thank you for posting! I wish I had found this video a day sooner..

KelsonHall-gz
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Thanks. I like the different approaches you took here. Really enlightening I would say. Also I've never actually seen someone evaluate for values of ln before. Cool video!

MathematicalToolbox