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OK I am completely lost on this log equation:
log(3-x) + log(3+x) = log5
Does anyone get an answer of ±2? If so HOW did you do it?
log(3-x) + log(3+x) = log5
Does anyone get an answer of ±2? If so HOW did you do it?
First of all, I think you expanded that wrong.ms. confused said:Exactly, so this is what I did:
log(x^2 - 9) = log5
(x^2 -14) = 0
My high school math teacher used to call it "dropping logs". I always found that funny.ShawnD said:Secondly, that's not how you "undo" a log.