Log Question Help

Scrutinize

Junior Member
Joined
Sep 16, 2019
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52
Solve for x:

3 log (x^2) − log^2 (−x) = 9

I've tried so many different ways to solve this but I can't get over the log^2(-x) because I've never learned what to do with a squared log. I just can't figure this one out.

Any help would be appreciated, thanks!
 
There's nothing special to learn about a squared log. Ignore it for the moment, and things will work out.

I would first simplify log(x^2) and log(-x), if possible, and then make a substitution (let u = ...) to transform the equation into a quadratic. That's where the square will come in.
 
I did try doing that, but what I ended up getting was 6(log(x)) - (log(x) + log(-1))^2 = 9. Problem with that is log(-1) is undefined and I can't really do anything else because its all inside the squared and has addition with it. Am I doing it wrong by doing it like that?
 
Yes, since log(-1) is undefined, you can't do that. That's why I said, "if possible". It's a discovery I wanted you to make. (By the way, if you followed our guidelines and showed your work first, we'd be past this step by now ...)

So, assume, as you must, that x is negative; and put everything in terms of log(-x) rather than log(x). That will be your u.
 
Apologies, just thought it was wrong. If I wrote each way I tried solving it here, it would've been much more confusing though because I tried like 10 different ways. Anyway I subsituted u in for log(-x) and I got 6log(x) - u^2 = 9, the problem is I'm not sure how I'm supposed to convert the log(x) into log(-x)?
Thanks for the help!
 
Attached is the work:

Thank you very much both of you!
 

Attachments

  • Log Problem.pdf
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Good.

And, as always, you can check your answer by substituting it in the original equation: 3 log ((-1000)^2) − log^2 (−(-1000)) = 3*6 - 3^2 = 9 .
 
Good.

And, as always, you can check your answer by substituting it in the original equation: 3 log ((-1000)^2) − log^2 (−(-1000)) = 3*6 - 3^2 = 9 .
Is your name just Dr.Peterson randomly, or do you actually have a Ph.D? Random question.
 
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