What is expand ln((5g^{-1})/w) ?

The solution to expand ln((5g^{-1})/w) is ln(5)-ln(g)-ln(w)

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expand ln((5g^{-1})/w)

expand

ln (\frac{5 g ^{- 1}}{w})

ln (5) - ln (g) - ln (w)

Solution steps

ln (\frac{5 g ^{- 1}}{w})

Apply log rule:

lo g_{c} (\frac{a}{b}) = lo g_{c} (a) - lo g_{c} (b)

ln (\frac{5 g ^{- 1}}{w}) = ln (5 g^{- 1}) - ln (w)

= ln (5 g^{- 1}) - ln (w)

Apply log rule:

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

ln (5 g^{- 1}) = ln (5) + ln (g^{- 1})

= ln (5) + ln (g^{- 1}) - ln (w)

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

ln (g^{- 1}) = - 1 \cdot ln (g)

= ln (5) - 1 \cdot ln (g) - ln (w)

Multiply:

1 \cdot ln (g) = ln (g)

= ln (5) - ln (g) - ln (w)

- lo g_{10} (3.55 \cdot 1 0^{- 32})

(ln (b + 2 a) - ln (a)) k - (a + 2 b) (a + b)

8 ln (x + 6) - 5 ln (x)

- lo g_{10} (2.17 \cdot 1 0^{- 6})

5 ln (x) + ln (x^{2} + 7) - 9 ln (x + 7)

What is expand ln((5g^{-1})/w) ?
The solution to expand ln((5g^{-1})/w) is ln(5)-ln(g)-ln(w)