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anonymous

  • 5 years ago

How do I solve log4y+log2y=12 ?

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  1. anonymous
    • 5 years ago
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    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

  2. anonymous
    • 5 years ago
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    hmm but my teacher's answer is 2 Wrong or Correct?

  3. anonymous
    • 5 years ago
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    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

  4. anonymous
    • 5 years ago
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    get the teacher to check the answer, or check the question again to be sure you transcribed it properly

  5. anonymous
    • 5 years ago
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    Thanks guys

  6. anonymous
    • 5 years ago
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    Could the teacher's equation be: \[\log(4y)+\log(2y)=\log(12)\] ?????

  7. anonymous
    • 5 years ago
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    Then maybe... \[\log(8y^2)=\log(12)\Rightarrow {\log}{8}+\log{y^2}=\log(12)\]

  8. anonymous
    • 5 years ago
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    Whoops, I thought I could get your teacher's answer with the above equation. I'll finish the solution anyway. \[\log(y^2)=\log(\frac{12}{8}) \] \[10^{\log(y^2)}=10^{({\log12/8})} \Rightarrow y^2=\frac{12}{8}\]

  9. anonymous
    • 5 years ago
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    \[y=\frac{\sqrt3}{\sqrt2}=\frac{\sqrt6}{2}\] -- 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").

  10. anonymous
    • 5 years ago
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    ya, I was assuming natural log, i.e. base e.

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