What is ln((a^{-2})/(b^1c^{-3)}) ?

The solution to ln((a^{-2})/(b^1c^{-3)}) is -2ln(a)-ln(b)+3ln(c)

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simplify ln((a^{-2})/(b^1c^{-3)})

simplify

ln (\frac{a ^{- 2}}{b ^{1} c ^{- 3}})

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

Solution steps

ln (\frac{a ^{- 2}}{b ^{1} c ^{- 3}})

Apply log rule:

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

ln (\frac{a ^{- 2}}{b ^{1} c ^{- 3}}) = ln (a^{- 2}) - ln (b^{1} c^{- 3})

= ln (a^{- 2}) - ln (b^{1} c^{- 3})

Apply log rule:

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

ln (a^{- 2}) = - 2 ln (a)

= - 2 ln (a) - ln (b^{1} c^{- 3})

Apply log rule:

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

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

= - 2 ln (a) - (ln (b^{1}) + ln (c^{- 3}))

Apply log rule:

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

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

= - 2 ln (a) - (1 \cdot ln (b) + ln (c^{- 3}))

Apply log rule:

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

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

= - 2 ln (a) - (1 \cdot ln (b) - 3 ln (c))

Multiply:

1 \cdot ln (b) = ln (b)

= - 2 ln (a) - (ln (b) - 3 ln (c))

- (ln (b) - 3 ln (c)) : - ln (b) + 3 ln (c)

= - 2 ln (a) - ln (b) + 3 ln (c)

lo g_{11} (\frac{x + y}{x ^{2} y})

ln (\frac{( 2 ^{(- 2)} )}{( 3 ^{(2)} 5 ^{(4)} )})

\frac{1}{4} ln (2 + e^{3 x}) - \frac{5}{12} ln (2) - 15 \frac{x}{12} + c

- lo g_{10} (0.1 M)

x + 4 lo g_{5} (y) - 2 lo g_{5} (z)

What is ln((a^{-2})/(b^1c^{-3)}) ?
The solution to ln((a^{-2})/(b^1c^{-3)}) is -2ln(a)-ln(b)+3ln(c)