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How could I find log4y+log2y=12?

Mathematics
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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 \[y=e^{6}/2\sqrt{2}\]
agreed
Thank alot aclandt, I was stuck in this question for more than 5hrs

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Other answers:

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

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