What is 3*(-2ln(11)+4ln(3))/(20) ?

The solution to 3*(-2ln(11)+4ln(3))/(20) is (3ln(81/121))/(20)

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simplify 3*(-2ln(11)+4ln(3))/(20)

simplify

3 \cdot \frac{- 2 ln ( 11 ) + 4 ln ( 3 )}{20}

\frac{3 ln ( \frac{81}{121} )}{20}

Decimal Notation

- 0.06020 \dots

Solution steps

3 \cdot \frac{- 2 ln ( 11 ) + 4 ln ( 3 )}{20}

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

2 ln (11) = ln (1 1^{2})

= 3 \cdot \frac{4 ln ( 3 ) - ln ( 1 1 ^{2} )}{20}

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

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

= 3 \cdot \frac{ln ( 3 ^{4} ) - ln ( 1 1 ^{2} )}{20}

Apply log rule:

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

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

= 3 \cdot \frac{ln ( \frac{3 ^{4}}{1 1 ^{2}} )}{20}

\frac{3 ^{4}}{1 1 ^{2}} = \frac{81}{121}

= 3 \cdot \frac{ln ( \frac{81}{121} )}{20}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{ln ( \frac{81}{121} ) \cdot 3}{20}

14 + lo g_{10} (\frac{0.118}{5})

ln (\frac{2 x}{1 + x ^{2}})

lo g_{4} ((\frac{12}{1 1 ^{3}})^{6})

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

- 0.38 \cdot lo g_{2} (0.38) - 0.62 lo g_{2} (0.62)

What is 3*(-2ln(11)+4ln(3))/(20) ?
The solution to 3*(-2ln(11)+4ln(3))/(20) is (3ln(81/121))/(20)