logarithm question

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\(\large \color{black}{\begin{align} \normalsize \text{Prove}\ \large \dfrac{3(\log_{10}242+\log_{10}45)+2\log_{10}80-5}{6}=\log_{10}66\hspace{.33em}\\~\\ \end{align}}\)
this is my thought \[\frac{\log(242^3)+\log(45^3)+\log(80^2)-\log(10^5)}{6} \\ =\frac{1}{6} \log(\frac{242^3 \cdot 45^3 \cdot 80^2}{10^5})\]
and then you can use power rule bring that 1/6 up and show \[(\frac{242^3 \cdot 45^3 \cdot 80^2}{10^5})^\frac{1}{6} = 66 \]

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thanks
\[242=2 \cdot 11^2 \\ 45 =9 \cdot 5=5 \cdot 3^2 \\ 80^2=2^4 \cdot 5 \\ 10=2 \cdot 5\\ \text{ so we have } \\ (\frac{2^3 \cdot 11^6 \cdot 5^3 \cdot 3^6 \cdot 2^{8} 5^2}{2^5 \cdot 5^5})^\frac{1}{6} \\ (2^{3+8-5}\cdot 11^6 \cdot 5^{3+2-5} \cdot 3^6)^\frac{1}{6}\]
so you don't have to use a calculator I chose prime factorization and law of exponents
as you will see this will work out nicely :)
yea i see the idea of prime factors

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