What is ln((a^{-3})/(b^1c^1)) ?

The solution to ln((a^{-3})/(b^1c^1)) is -3ln(a)-ln(b)-ln(c)

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simplify ln((a^{-3})/(b^1c^1))

simplify

ln (\frac{a ^{- 3}}{b ^{1} c ^{1}})

- 3 ln (a) - ln (b) - ln (c)

Solution steps

ln (\frac{a ^{- 3}}{b ^{1} c ^{1}})

Apply log rule:

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

ln (\frac{a ^{- 3}}{b ^{1} c ^{1}}) = ln (a^{- 3}) - ln (b^{1} c^{1})

= ln (a^{- 3}) - ln (b^{1} c^{1})

Apply log rule:

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

ln (a^{- 3}) = - 3 ln (a)

= - 3 ln (a) - ln (b^{1} c^{1})

Apply log rule:

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

ln (b^{1} c^{1}) = ln (b^{1}) + ln (c^{1})

= - 3 ln (a) - (ln (b^{1}) + ln (c^{1}))

Apply log rule:

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

ln (b^{1}) = 1 \cdot ln (b)

= - 3 ln (a) - (1 \cdot ln (b) + ln (c^{1}))

Apply log rule:

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

ln (c^{1}) = 1 \cdot ln (c)

= - 3 ln (a) - (1 \cdot ln (b) + 1 \cdot ln (c))

- (1 \cdot ln (b) + 1 \cdot ln (c)) = - (ln (b) + ln (c))

= - 3 ln (a) - (ln (b) + ln (c))

- (ln (b) + ln (c)) : - ln (b) - ln (c)

= - 3 ln (a) - ln (b) - ln (c)

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What is ln((a^{-3})/(b^1c^1)) ?
The solution to ln((a^{-3})/(b^1c^1)) is -3ln(a)-ln(b)-ln(c)