What is ln((2^{-3})/(3^{-4)5^{-2}}) ?

The solution to ln((2^{-3})/(3^{-4)5^{-2}}) is -3ln(2)+4ln(3)+2ln(5)

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simplify ln((2^{-3})/(3^{-4)5^{-2}})

simplify

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

- 3 ln (2) + 4 ln (3) + 2 ln (5)

Decimal Notation

5.53388 \dots

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

ln (2^{- 3}) = - 3 ln (2)

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

Apply log rule:

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

ln (3^{- 4} \cdot 5^{- 2}) = ln (3^{- 4}) + ln (5^{- 2})

= - 3 ln (2) - (ln (3^{- 4}) + ln (5^{- 2}))

Apply log rule:

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

ln (3^{- 4}) = - 4 ln (3)

= - 3 ln (2) - (- 4 ln (3) + ln (5^{- 2}))

Apply log rule:

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

ln (5^{- 2}) = - 2 ln (5)

= - 3 ln (2) - (- 4 ln (3) - 2 ln (5))

- (- 4 ln (3) - 2 ln (5)) : 4 ln (3) + 2 ln (5)

= - 3 ln (2) + 4 ln (3) + 2 ln (5)

ln (\frac{30.3}{10.6})

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ln (\frac{( λ ^{(x_{1})} e ^{(- λ)} )}{( x _{1} ! )})

What is ln((2^{-3})/(3^{-4)5^{-2}}) ?
The solution to ln((2^{-3})/(3^{-4)5^{-2}}) is -3ln(2)+4ln(3)+2ln(5)