What is log_{7}(1/7 mn^2) ?

The solution to log_{7}(1/7 mn^2) is -1+log_{7}(m)+2log_{7}(n)

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simplify log_{7}(1/7 mn^2)

simplify

lo g_{7} (\frac{1}{7} m n^{2})

- 1 + lo g_{7} (m) + 2 lo g_{7} (n)

Solution steps

lo g_{7} (\frac{1}{7} m n^{2})

\frac{1}{7} = 7^{- 1}

= lo g_{7} (7^{- 1} m n^{2})

Apply log rule:

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

lo g_{7} (7^{- 1} m n^{2}) = lo g_{7} (7^{- 1}) + lo g_{7} (m n^{2})

= lo g_{7} (7^{- 1}) + lo g_{7} (m n^{2})

Apply log rule:

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

lo g_{7} (7^{- 1}) = - 1

= - 1 + lo g_{7} (m n^{2})

Simplify

lo g_{7} (m n^{2}) : lo g_{7} (m) + 2 lo g_{7} (n)

= - 1 + (lo g_{7} (m) + 2 lo g_{7} (n))

Apply rule:

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

- 1 + (lo g_{7} (m) + 2 lo g_{7} (n)) = - 1 + lo g_{7} (m) + 2 lo g_{7} (n)

= - 1 + lo g_{7} (m) + 2 lo g_{7} (n)

- ln ∣ v + 1 ∣ - \frac{1}{2} - ln v^{2} 11

lo g_{9} (\frac{x}{y})

4 (\frac{1}{9} ln ∣ x - 2 ∣ - \frac{1}{9} ln ∣ x + 7 ∣ + C)

ln (\frac{1}{2}) + ln (\frac{2}{3})

lo g_{10} (2) - lo g_{10} (1.035)

What is log_{7}(1/7 mn^2) ?
The solution to log_{7}(1/7 mn^2) is -1+log_{7}(m)+2log_{7}(n)