What is ln((2^{-4})/(3^{-3)5^{-1}}) ?

The solution to ln((2^{-4})/(3^{-3)5^{-1}}) is -4ln(2)+3ln(3)+ln(5)

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simplify ln((2^{-4})/(3^{-3)5^{-1}})

simplify

ln (\frac{2 ^{- 4}}{3 ^{- 3} 5 ^{- 1}})

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

Decimal Notation

2.13268 \dots

Solution steps

ln (\frac{2 ^{- 4}}{3 ^{- 3} \cdot 5 ^{- 1}})

Apply log rule:

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

ln (\frac{2 ^{- 4}}{3 ^{- 3} \cdot 5 ^{- 1}}) = ln (2^{- 4}) - ln (3^{- 3} \cdot 5^{- 1})

= ln (2^{- 4}) - ln (5^{- 1} \cdot 3^{- 3})

Apply log rule:

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

ln (2^{- 4}) = - 4 ln (2)

= - 4 ln (2) - ln (3^{- 3} \cdot 5^{- 1})

Apply log rule:

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

ln (3^{- 3} \cdot 5^{- 1}) = ln (3^{- 3}) + ln (5^{- 1})

= - 4 ln (2) - (ln (3^{- 3}) + ln (5^{- 1}))

Apply log rule:

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

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

= - 4 ln (2) - (- 3 ln (3) + ln (5^{- 1}))

Apply log rule:

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

ln (5^{- 1}) = - 1 \cdot ln (5)

= - 4 ln (2) - (- 3 ln (3) - 1 \cdot ln (5))

Multiply:

1 \cdot ln (5) = ln (5)

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

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

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

ln (\frac{78}{175}) 6

lo g_{10} (\frac{8 z}{11})

\frac{1}{2} lo g_{10} (a) + \frac{1}{2} lo g_{10} (b)

- lo g_{10} (4.5 \times 1 0^{- 4})

\frac{ln ( \frac{8}{9} )}{2 ln ( 6 )}

What is ln((2^{-4})/(3^{-3)5^{-1}}) ?
The solution to ln((2^{-4})/(3^{-3)5^{-1}}) is -4ln(2)+3ln(3)+ln(5)