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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

simplify

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

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

Decimal Notation

4.76377 \dots

Solution steps

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

Apply log rule:

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

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

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

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^{- 1} \cdot 5^{- 4})

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Apply log rule:

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

ln (5^{- 4}) = - 4 ln (5)

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

Multiply:

1 \cdot ln (3) = ln (3)

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

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

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

ln (x^{2}) + 2 ln (x)

3 ln (x) - \frac{1}{4} ln (y)

lo g_{4} (5) + lo g_{4} (x) + lo g_{4} (y)

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