anonymous
  • anonymous
How could I find log4y+log2y=12?
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
rasing e to the relative powers gives \[e ^{\log(4y)+\log(2y)}=e ^{12}\] seperating the powers of e then gives \[e ^{\log(4y)}e ^{\log(2y)} = e ^{12}\] hence \[8y ^{2} = e ^{12}\] so solving from there gives \[y=e^{6}/2\sqrt{2}\]
anonymous
  • anonymous
agreed
anonymous
  • anonymous
Thank alot aclandt, I was stuck in this question for more than 5hrs

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anonymous
  • anonymous
assuming log with base e
anonymous
  • anonymous
yeah, though unless given otherwise you should always assume log to be to base e

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