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

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

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

simplify

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

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

Decimal Notation

- 4.49980 \dots

Solution steps

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

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Multiply:

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

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

Multiply:

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

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

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

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

ln (9) + 2 ln (4) - ln (2)

\frac{ln ( 29500 ) - ln ( 100 )}{ln ( 75000 ) - ln ( 100 )}

ln (\frac{1.5}{11})

ln (\frac{x ^{2} - x - 2}{( x + 4 ) ^{2}})

\frac{600 ln ( 2 )}{3 ln ( 2 ) - ln ( 3 )}

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