What is ln|-(mg)/b+a|-ln((|m||g|)/(|b|)) ?

The solution to ln|-(mg)/b+a|-ln((|m||g|)/(|b|)) is ln((|-mg+ba|)/(|m||g|))

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simplify ln|-(mg)/b+a|-ln((|m||g|)/(|b|))

simplify

ln - \frac{m g}{b} + a - ln (\frac{∣ m ∣ ∣ g ∣}{∣ b ∣})

ln (\frac{∣ - m g + ba ∣}{∣ m ∣ ∣ g ∣})

Solution steps

ln - \frac{m g}{b} + a - ln (\frac{∣ m ∣ ∣ g ∣}{∣ b ∣})

Apply log rule:

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

ln - \frac{m g}{b} + a - ln (\frac{∣ m ∣ ∣ g ∣}{∣ b ∣}) = ln \frac{a - \frac{m g}{b}}{\frac{∣ m ∣ ∣ g ∣}{∣ b ∣}}

= ln \frac{- \frac{m g}{b} + a}{\frac{∣ m ∣ ∣ g ∣}{∣ b ∣}}

Simplify

\frac{- \frac{m g}{b} + a}{\frac{∣ m ∣ ∣ g ∣}{∣ b ∣}} : \frac{∣ - m g + ba ∣}{∣ m ∣ ∣ g ∣}

= ln (\frac{∣ - m g + ba ∣}{∣ m ∣ ∣ g ∣})

- lo g_{10} (2.92 \cdot 1 0^{- 14})

\frac{3}{2} ln ∣ (- 0.5 - 1) ∣ - \frac{3}{2} ln ∣ (- 1 - 1) ∣

ln (x^{3} \cdot (x + 2))

\frac{2 cos ( 2 ln ( x ) )}{x ^{2}} + x^{3}

2 ln (a) - \frac{1}{2} ln (b)

What is ln|-(mg)/b+a|-ln((|m||g|)/(|b|)) ?
The solution to ln|-(mg)/b+a|-ln((|m||g|)/(|b|)) is ln((|-mg+ba|)/(|m||g|))