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mathmath333

  • one year ago

logarithm question

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} \normalsize \text{Prove}\ \large \dfrac{3(\log_{10}242+\log_{10}45)+2\log_{10}80-5}{6}=\log_{10}66\hspace{.33em}\\~\\ \end{align}}\)

  2. freckles
    • one year ago
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    this is my thought \[\frac{\log(242^3)+\log(45^3)+\log(80^2)-\log(10^5)}{6} \\ =\frac{1}{6} \log(\frac{242^3 \cdot 45^3 \cdot 80^2}{10^5})\]

  3. freckles
    • one year ago
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    and then you can use power rule bring that 1/6 up and show \[(\frac{242^3 \cdot 45^3 \cdot 80^2}{10^5})^\frac{1}{6} = 66 \]

  4. mathmath333
    • one year ago
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    thanks

  5. freckles
    • one year ago
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    \[242=2 \cdot 11^2 \\ 45 =9 \cdot 5=5 \cdot 3^2 \\ 80^2=2^4 \cdot 5 \\ 10=2 \cdot 5\\ \text{ so we have } \\ (\frac{2^3 \cdot 11^6 \cdot 5^3 \cdot 3^6 \cdot 2^{8} 5^2}{2^5 \cdot 5^5})^\frac{1}{6} \\ (2^{3+8-5}\cdot 11^6 \cdot 5^{3+2-5} \cdot 3^6)^\frac{1}{6}\]

  6. freckles
    • one year ago
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    so you don't have to use a calculator I chose prime factorization and law of exponents

  7. freckles
    • one year ago
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    as you will see this will work out nicely :)

  8. mathmath333
    • one year ago
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    yea i see the idea of prime factors

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