What is ln((2^{-1})/(3^45^1)) ?

The solution to ln((2^{-1})/(3^45^1)) is -ln(2)-4ln(3)-ln(5)

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simplify ln((2^{-1})/(3^45^1))

simplify

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

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

Decimal Notation

- 6.69703 \dots

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

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

Multiply:

1 \cdot ln (2) = ln (2)

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

Multiply:

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

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

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

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

- lo g_{10} (1.3 \cdot 1 0^{- 12})

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

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

ln (\frac{1.6}{0.277})

ln (x y) - cos (x) d x

What is ln((2^{-1})/(3^45^1)) ?
The solution to ln((2^{-1})/(3^45^1)) is -ln(2)-4ln(3)-ln(5)