What is log_{10}(1/(0.4k)) ?

The solution to log_{10}(1/(0.4k)) is -2log_{10}(2)-log_{10}(k)+1

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simplify log_{10}(1/(0.4k))

simplify

lo g_{10} (\frac{1}{0.4 k})

- 2 lo g_{10} (2) - lo g_{10} (k) + 1

Solution steps

lo g_{10} (\frac{1}{0.4 k})

Simplify

\frac{1}{0.4 k} : (0.4 k)^{- 1}

= lo g_{10} ((0.4 k)^{- 1})

Apply log rule

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

lo g_{10} ((0.4 k)^{- 1}) = - 1 \cdot lo g_{10} (0.4 k)

= - 1 \cdot lo g_{10} (0.4 k)

Apply rule:

1 \cdot a = a

= - lo g_{10} (0.4 k)

Apply log rule:

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

- lo g_{10} (0.4 k) = - (lo g_{10} (0.4) + lo g_{10} (k))

= - (lo g_{10} (0.4) + lo g_{10} (k))

lo g_{10} (0.4) = 2 lo g_{10} (2) - 1

= - (2 lo g_{10} (2) - 1 + lo g_{10} (k))

Apply the distributive law:

- (a - b) = - a + b

- (2 lo g_{10} (2) - 1 + lo g_{10} (k)) = - 2 lo g_{10} (2) + 1 - lo g_{10} (k)

= - 2 lo g_{10} (2) + 1 - lo g_{10} (k)

Apply log rule:

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

2 lo g_{10} (2) = lo g_{10} (2^{2})

= - lo g_{10} (2^{2}) + 1 - lo g_{10} (k)

Rewrite

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

- (lo g_{10} (2^{2}) + lo g_{10} (k))

= - (lo g_{10} (k) + lo g_{10} (2^{2})) + 1

Apply log rule:

lo g_{c} (a) + lo g_{c} (b) = lo g_{c} (ab)

lo g_{10} (2^{2}) + lo g_{10} (k) = lo g_{10} (2^{2} k)

= - lo g_{10} (2^{2} k) + 1

2^{2} = 4

= - lo g_{10} (4 k) + 1

Simplify

- lo g_{10} (4 k) : - 2 lo g_{10} (2) - lo g_{10} (k)

= - 2 lo g_{10} (2) - lo g_{10} (k) + 1

ln (12) + ln (60)

ln (- 2400 e^{- 300 t})

- ln (\frac{k ^{2}}{k ^{1}})

- lo g_{10} (5.8 \cdot 1 0^{- 8})

ln (8 \cdot e^{- 8 x})

What is log_{10}(1/(0.4k)) ?
The solution to log_{10}(1/(0.4k)) is -2log_{10}(2)-log_{10}(k)+1