Jamierox4ev3r
  • Jamierox4ev3r
Rewrite the following log expression using the law of logarithms. Your final answer should have no exponents and many logs.
Mathematics
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SOLVED
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chestercat
  • chestercat
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Jamierox4ev3r
  • Jamierox4ev3r
\(\huge\log \left( \frac{ x^{7} }{ \sqrt[3]{y^{2}z^{5}} } \right)\)
Jamierox4ev3r
  • Jamierox4ev3r
I believe I have this one figured out, but I'm not sure, so I'll post my work and final answer
Jamierox4ev3r
  • Jamierox4ev3r
\(\huge\log \left( \frac{ x^{7} }{ \sqrt[3]{y^{2}z^{5}} } \right)\) \(\Large\log x^{7} - \log \sqrt[3]{y^{2}z^{5}}\) \(\Large 7\log x - \log (y^{2}z^{5})^{\frac{1}{3}}\) \(\Large 7\log x - \log (y^{\frac{2}{3}} z^{\frac{5}{3}})\) \(\Large 7\log x - \log y^{\frac{2}{3}} + \log z^{\frac{5}{3}}\) \(\Huge 7\log x - \frac{2}{3}\log y + \frac{5}{3}\log z\) ^^ 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.

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More answers

anonymous
  • anonymous
there's mistake in one sign.. check it..
Jamierox4ev3r
  • Jamierox4ev3r
sign? you mean a discrepancy for whether something is positive or negative?
anonymous
  • anonymous
yes
Jamierox4ev3r
  • Jamierox4ev3r
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.
anonymous
  • anonymous
you stated right but check this -( log MN) = -log M -log N if you agree then check again that step
Jamierox4ev3r
  • Jamierox4ev3r
Oh. So you believe that the negative sign distributes into the parentheses?
anonymous
  • anonymous
I dont believe.. its maths :P
Jamierox4ev3r
  • Jamierox4ev3r
lol whoops, wrong way of phrasing that. But I see what you mean, and thanks
anonymous
  • anonymous
:)
Jamierox4ev3r
  • Jamierox4ev3r
\(\Huge 7\log x - \frac{2}{3}\log y - \frac{5}{3}\log z\) ^^ That should be the final answer then, if I'm not mistaken
anonymous
  • anonymous
yep
Jamierox4ev3r
  • Jamierox4ev3r
Awesome, I get it. Thank you ^_^
anonymous
  • anonymous
nvm :)

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