What is log_{10}((1/9)/k) ?

The solution to log_{10}((1/9)/k) is -2log_{10}(3)-log_{10}(k)

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simplify log_{10}((1/9)/k)

simplify

lo g_{10} (\frac{\frac{1}{9}}{k})

- 2 lo g_{10} (3) - lo g_{10} (k)

Solution steps

lo g_{10} (\frac{\frac{1}{9}}{k})

Simplify

\frac{\frac{1}{9}}{k} : \frac{1}{9 k}

= lo g_{10} (\frac{1}{9 k})

Simplify

\frac{1}{9 k} : (9 k)^{- 1}

= lo g_{10} ((9 k)^{- 1})

Apply log rule

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x), assuming x > 0

lo g_{10} ((9 k)^{- 1}) = - 1 \cdot lo g_{10} (9 k)

= - 1 \cdot lo g_{10} (9 k)

Apply rule:

1 \cdot a = a

= - lo g_{10} (9 k)

Apply log rule:

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

- lo g_{10} (9 k) = - (lo g_{10} (9) + lo g_{10} (k))

= - (lo g_{10} (9) + lo g_{10} (k))

lo g_{10} (9) = 2 lo g_{10} (3)

= - (2 lo g_{10} (3) + lo g_{10} (k))

Apply the distributive law:

- (a + b) = - a - b

- (2 lo g_{10} (3) + lo g_{10} (k)) = - 2 lo g_{10} (3) - lo g_{10} (k)

= - 2 lo g_{10} (3) - lo g_{10} (k)

ln (x y) + 9 ln (\frac{y}{z}) - 8 ln (x yz)

5 lo g_{10} (y) - 3 lo g_{10} (m)

5 lo g_{6} (u) + 8 lo g_{6} (v)

- lo g_{10} (4 \cdot 1 0^{- 8}) + lo g_{10} (1)

2.338 - 1.5 - 0.287 ln (\frac{675}{45})

What is log_{10}((1/9)/k) ?
The solution to log_{10}((1/9)/k) is -2log_{10}(3)-log_{10}(k)