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- Jamierox4ev3r

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

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- Jamierox4ev3r

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- Jamierox4ev3r

\(\huge\log \left( \frac{ x^{7} }{ \sqrt[3]{y^{2}z^{5}} } \right)\)

- Jamierox4ev3r

I believe I have this one figured out, but I'm not sure, so I'll post my work and final answer

- 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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- anonymous

there's mistake in one sign.. check it..

- Jamierox4ev3r

sign? you mean a discrepancy for whether something is positive or negative?

- anonymous

yes

- 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

you stated right but check this -( log MN) = -log M -log N
if you agree then check again that step

- Jamierox4ev3r

Oh. So you believe that the negative sign distributes into the parentheses?

- anonymous

I dont believe.. its maths :P

- Jamierox4ev3r

lol whoops, wrong way of phrasing that. But I see what you mean, and thanks

- anonymous

:)

- 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

yep

- Jamierox4ev3r

Awesome, I get it. Thank you ^_^

- anonymous

nvm :)

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