anonymous
  • anonymous
How do I solve log4y+log2y=12 ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
log 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
  • anonymous
hmm but my teacher's answer is 2 Wrong or Correct?
anonymous
  • anonymous
I 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

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anonymous
  • anonymous
get the teacher to check the answer, or check the question again to be sure you transcribed it properly
anonymous
  • anonymous
Thanks guys
anonymous
  • anonymous
Could the teacher's equation be: \[\log(4y)+\log(2y)=\log(12)\] ?????
anonymous
  • anonymous
Then maybe... \[\log(8y^2)=\log(12)\Rightarrow {\log}{8}+\log{y^2}=\log(12)\]
anonymous
  • anonymous
Whoops, 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
  • anonymous
\[y=\frac{\sqrt3}{\sqrt2}=\frac{\sqrt6}{2}\] -- 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
  • anonymous
ya, I was assuming natural log, i.e. base e.

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