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

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

simplify

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

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

Decimal Notation

6.15006 \dots

Solution steps

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

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

Apply log rule:

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

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

= - 2 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})

= - 2 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)

= - 2 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)

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

Multiply:

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

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

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

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

3 ln (\frac{1}{2})

lo g_{10} (\frac{1}{5} (\frac{a ^{2} b}{c}))

lo g_{10} (x) + \frac{1}{2} lo g_{10} (y) - 3 lo g_{10} (z)

2400 \cdot ln (\frac{750}{750 - 14 ( t )}) - (9.81 \cdot t)

5 lo g_{6} (u) + 2 lo g_{6} (v)

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