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explain how to evaluate log^6 64
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\( \color{Black}{\Rightarrow \log_6 64}\)?
yes
That wouldn't be a whole number, so I am thinking at the moment.
\( \color{Black}{\Rightarrow \Large {\log64 \over \log6} }\) And now you use a calculator.
You can simplify log 64 if you want as 2 log 8
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Yep. Let me explain how you get \(2 \log8\). \( \color{Black}{\Rightarrow \log(8)^2 }\) You bring 2 to the left according to a property. \( \color{Black}{\Rightarrow 2 \log 8}\)
Two steps heh
\[\large{\log(k^2)=2\log{(k)}}\]
okay can yu guys help me on my second problem now
\(\log_ab^c = c \log_ab\) Better :)
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