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OpenStudy (anonymous):

How could I find log4y+log2y=12?

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OpenStudy (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 \[y=e^{6}/2\sqrt{2}\]

OpenStudy (anonymous):

agreed

OpenStudy (anonymous):

Thank alot aclandt, I was stuck in this question for more than 5hrs

OpenStudy (anonymous):

assuming log with base e

OpenStudy (anonymous):

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

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