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Jamierox4ev3r

  • one year ago

Rewrite the following log expression using the law of logarithms. Your final answer should have no exponents and many logs.

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  1. Jamierox4ev3r
    • one year ago
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    \(\huge\log \left( \frac{ x^{7} }{ \sqrt[3]{y^{2}z^{5}} } \right)\)

  2. Jamierox4ev3r
    • one year ago
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    I believe I have this one figured out, but I'm not sure, so I'll post my work and final answer

  3. Jamierox4ev3r
    • one year ago
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    \(\huge\log \left( \frac{ x^{7} }{ \sqrt[3]{y^{2}z^{5}} } \right)\) \(\Large\log x^{7} - \log \sqrt[3]{y^{2}z^{5}}\) \(\Large 7\log x - \log (y^{2}z^{5})^{\frac{1}{3}}\) \(\Large 7\log x - \log (y^{\frac{2}{3}} z^{\frac{5}{3}})\) \(\Large 7\log x - \log y^{\frac{2}{3}} + \log z^{\frac{5}{3}}\) \(\Huge 7\log x - \frac{2}{3}\log y + \frac{5}{3}\log z\) ^^ I believe that is my final answer. Someone please let me know if I'm right, or if I'm not, tell me where I went wrong. Thanks in advance.

  4. anonymous
    • one year ago
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    there's mistake in one sign.. check it..

  5. Jamierox4ev3r
    • one year ago
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    sign? you mean a discrepancy for whether something is positive or negative?

  6. anonymous
    • one year ago
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    yes

  7. Jamierox4ev3r
    • one year ago
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    Hmm...I think you may be talking about the 4/5 step, where I expanded the parentheses. but isn't it a rule that states this? log(MN) = logM + logN? Therefore, I don't see any error in my process there; at least, I don't think I see an error there.

  8. anonymous
    • one year ago
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    you stated right but check this -( log MN) = -log M -log N if you agree then check again that step

  9. Jamierox4ev3r
    • one year ago
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    Oh. So you believe that the negative sign distributes into the parentheses?

  10. anonymous
    • one year ago
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    I dont believe.. its maths :P

  11. Jamierox4ev3r
    • one year ago
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    lol whoops, wrong way of phrasing that. But I see what you mean, and thanks

  12. anonymous
    • one year ago
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    :)

  13. Jamierox4ev3r
    • one year ago
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    \(\Huge 7\log x - \frac{2}{3}\log y - \frac{5}{3}\log z\) ^^ That should be the final answer then, if I'm not mistaken

  14. anonymous
    • one year ago
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    yep

  15. Jamierox4ev3r
    • one year ago
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    Awesome, I get it. Thank you ^_^

  16. anonymous
    • one year ago
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    nvm :)

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