What is log_{3}(1/9 mn^2) ?

The solution to log_{3}(1/9 mn^2) is -2+log_{3}(m)+2log_{3}(n)

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simplify log_{3}(1/9 mn^2)

simplify

lo g_{3} (\frac{1}{9} m n^{2})

- 2 + lo g_{3} (m) + 2 lo g_{3} (n)

Solution steps

lo g_{3} (\frac{1}{9} m n^{2})

\frac{1}{9} = 3^{- 2}

= lo g_{3} (3^{- 2} m n^{2})

Apply log rule:

lo g_{a} (x y) = lo g_{a} (x) + lo g_{a} (y)

lo g_{3} (3^{- 2} m n^{2}) = lo g_{3} (3^{- 2}) + lo g_{3} (m n^{2})

= lo g_{3} (3^{- 2}) + lo g_{3} (m n^{2})

Apply log rule:

lo g_{a} (a^{b}) = b

lo g_{3} (3^{- 2}) = - 2

= - 2 + lo g_{3} (m n^{2})

Simplify

lo g_{3} (m n^{2}) : lo g_{3} (m) + 2 lo g_{3} (n)

= - 2 + (lo g_{3} (m) + 2 lo g_{3} (n))

Apply rule:

a + (b + c) = a + b + c

- 2 + (lo g_{3} (m) + 2 lo g_{3} (n)) = - 2 + lo g_{3} (m) + 2 lo g_{3} (n)

= - 2 + lo g_{3} (m) + 2 lo g_{3} (n)

ln (\frac{x ^{4} y}{7})

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lo g_{10} (\frac{x ^{2}}{x})

lo g_{4} (\frac{y}{16})

\frac{1}{2} lo g_{c} (x) - 5 lo g_{c} (x) + 2 lo g_{c} (y)

What is log_{3}(1/9 mn^2) ?
The solution to log_{3}(1/9 mn^2) is -2+log_{3}(m)+2log_{3}(n)