What is 1/6*ln(1/(9^3)) ?

The solution to 1/6*ln(1/(9^3)) is -ln(3)

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simplify 1/6*ln(1/(9^3))

simplify

\frac{1}{6} \cdot ln (\frac{1}{9 ^{3}})

- ln (3)

Decimal Notation

- 1.09861 \dots

Solution steps

\frac{1}{6} ln (\frac{1}{9 ^{3}})

Apply log rule:

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

ln (\frac{1}{9 ^{3}}) = ln (1) - ln (9^{3})

= \frac{1}{6} (ln (1) - ln (9^{3}))

Apply log rule:

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

ln (9^{3}) = 3 ln (9)

= \frac{1}{6} (ln (1) - 3 ln (9))

Apply log rule:

lo g_{a} (1) = 0

= \frac{1}{6} (- 3 ln (9) + 0)

Multiply fractions:

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

= \frac{1 \cdot ( 0 - 3 ln ( 9 ) )}{6}

1 \cdot (0 - 3 ln (9)) = - 3 ln (9)

= \frac{- 3 ln ( 9 )}{6}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{3 ln ( 9 )}{6}

Cancel the common factor:

3

= - \frac{ln ( 9 )}{2}

Simplify

ln (9) : 2 ln (3)

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

Divide the numbers:

\frac{2}{2} = 1

= - ln (3)

lo g_{10} (z y^{3})

\frac{1}{2} [ln (3) + 4 ln (x) - 2 ln (x - 1)]

\frac{( 0.5 ln ( 28 ) - ln ( 2 ) )}{0.5}

\frac{1}{2} ln (\frac{11}{3})

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