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anonymous

  • 5 years ago

How could I find log4y+log2y=12?

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  1. anonymous
    • 5 years ago
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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 \[y=e^{6}/2\sqrt{2}\]

  2. anonymous
    • 5 years ago
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    agreed

  3. anonymous
    • 5 years ago
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    Thank alot aclandt, I was stuck in this question for more than 5hrs

  4. anonymous
    • 5 years ago
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    assuming log with base e

  5. anonymous
    • 5 years ago
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    yeah, though unless given otherwise you should always assume log to be to base e

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