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Mathematics
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Condensing Logarithms
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the equation is number 20 http://www.luminpdf.com/files/1476787/Unit%207%20Summer%20Work.pdf
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Same denominator, add the numerators first.
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\[\log_{9}uvw \] ?
over 3
That's right. Now, you have\[\large \dfrac{\log_9 uvw}{3} = \dfrac{\log_9 uvw}{\log_9 {9^3}}= \log_{9^3}{uvw}\]Do you follow?
thanks
that answer isn't one of the options
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Never mind that one. I have a better method.\[\large\dfrac{ \log_9{uvw}}{3} = \dfrac{1}{3}\log_9(uvw) = \log_9{\left(\left(uvw\right)^{1/3}\right)}\]
so the answer is \[\log_{9} \sqrt[3]{wvu}\]
indeed.
Thank you
No problem.
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