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anonymous
 5 years ago
How do I solve log4y+log2y=12 ?
anonymous
 5 years ago
How do I solve log4y+log2y=12 ?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0log 4y + log 2y = 12 => log ( 4y * 2y ) = 12 since log(A * B) = log A + log B => log 8y^2 = 12 => 8y^2 = (10)^12 => y^2 = (10)^12 / 8 => y = 10^6 / 8^1/2 therefore, y = 1000000 / 8^1/2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm but my teacher's answer is 2 Wrong or Correct?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have to go with aclandt's answer. It is my understanding that in the 3rd line of this answer, you have to use e on both sides to get the 8y^2 down, which gives you 8y^2 = e^12 which is the same as aclandt's

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0get the teacher to check the answer, or check the question again to be sure you transcribed it properly

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Could the teacher's equation be: \[\log(4y)+\log(2y)=\log(12)\] ?????

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then maybe... \[\log(8y^2)=\log(12)\Rightarrow {\log}{8}+\log{y^2}=\log(12)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Whoops, I thought I could get your teacher's answer with the above equation. I'll finish the solution anyway. \[\log(y^2)=\log(\frac{12}{8}) \] \[10^{\log(y^2)}=10^{({\log12/8})} \Rightarrow y^2=\frac{12}{8}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[y=\frac{\sqrt3}{\sqrt2}=\frac{\sqrt6}{2}\]  So I didn't get 2 either. BTW, ulthi is right in using the base 10 to cancel the base 10 log. Raising with [\e\] is only to invert ("undo") the natural logs (written "ln").

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ya, I was assuming natural log, i.e. base e.
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