Phi for All Posted June 6, 2008 Posted June 6, 2008 Listen, tomas, you're going to need to give us a bit more background on this. Is this for homework? We don't do your homework for you. We can help aim you in the right direction but you need to talk to us here. What's going on?
tomas Posted June 6, 2008 Author Posted June 6, 2008 these is from another forum and i dont no how i can solve it thanks
Phi for All Posted June 6, 2008 Posted June 6, 2008 OK, thanks for the reply. Is it for school homework?
tomas Posted June 6, 2008 Author Posted June 6, 2008 no these is so defeclt , i thenk it is not school homework
Riogho Posted June 6, 2008 Posted June 6, 2008 I've been applying my n00b skills to this problem and I'm stumped. I'd be curious as to the answer as well. More importantly however, how you got it. I just keep getting myself caught in circles.
Kyrisch Posted June 7, 2008 Posted June 7, 2008 (edited) Well [math] 9 = 3^2 [/math] and your log bases are 3... There's a clue . Also, I'm having trouble putting in the log base subscript with LaTeX... how do you do that? Edited June 7, 2008 by Kyrisch
Bignose Posted June 7, 2008 Posted June 7, 2008 Well, there aren't any restrictions on how to solve it. So I graphed it. There is an answer. I don't know if there are any pure analytical ways to get to it -- but numerical solutions are just as valid as any other method.
Kyrisch Posted June 7, 2008 Posted June 7, 2008 [math] \log_3{(4x^2-9)}^{\log_3{(x+6)}}=3^2[/math] [math] 3^{3^2} = (4x^2 - 9)^{\log_3{(x+6)}} [/math] [math] \log_{4x^2-9}{3^9} = \log_3{(x+6)} [/math] [math] 9\log_{4x^2-9}{3} = \log_3{(x+6)} [/math] [math] 9 \frac{\log_3{3}}{\log_3{(4x^2-9)}} = \log_3{(x+6)} [/math] [math] \frac{9}{\log_3{(4x^2-9)}} = \log_3{(x+6)} [/math] ...is as far as I can get
Riogho Posted June 7, 2008 Posted June 7, 2008 I got it all on one line, but my skills are really rusty. Forgive me, I don't know LateX either. log(x+6) log(2log(4x-9)) = log(9)log(3)log(3) That's as far as I got
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