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Precalculus
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OpenStudy (anonymous):
Plzz help me How would you write the following expression as a sum or difference? log(3√2-x/3x) A.1/3(log(2-x)+log3x) B. 1/3log(2-x)-log3x C. 1/3log(2-x)-logx D. 1/3log(2-x)+log3x
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OpenStudy (anonymous):
this is a review for a test
OpenStudy (anonymous):
\[\log\left(\frac{\sqrt[3]{2-x}}{3x}\right)\]?
OpenStudy (anonymous):
yes exactly
OpenStudy (anonymous):
or \[\log\left(\sqrt[3]{\frac{2-x}{3x}}\right)\]
OpenStudy (anonymous):
how did you write it like that
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OpenStudy (anonymous):
magic
OpenStudy (anonymous):
the first one
OpenStudy (anonymous):
then start with \[\log(\sqrt[3]{2-x})-\log(3x)\]
OpenStudy (anonymous):
then go to \[\frac{1}{3}\log(2-x)-\log(3x)\]
OpenStudy (anonymous):
Ahh i see
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OpenStudy (anonymous):
do u mind helping me with one more question?
OpenStudy (anonymous):
sure and btw i am writing it in \[\LaTeX\]
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