What is 1/2 ln((1+(sqrt(2))/2)/(1-(sqrt(2))/2)) ?

The solution to 1/2 ln((1+(sqrt(2))/2)/(1-(sqrt(2))/2)) is (ln(1+(sqrt(2))/2)-ln(1-(sqrt(2))/2))/(2)

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simplify 1/2 ln((1+(sqrt(2))/2)/(1-(sqrt(2))/2))

simplify

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

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

Decimal Notation

0.88137 \dots

Solution steps

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

Apply log rule:

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

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

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

Multiply fractions:

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

= \frac{1 \cdot ( ln ( 1 + \frac{2}{2} ) - ln ( 1 - \frac{2}{2} ) )}{2}

1 \cdot (ln (1 + \frac{2}{2}) - ln (1 - \frac{2}{2})) = ln (1 + \frac{2}{2}) - ln (1 - \frac{2}{2})

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

Cancel

\frac{2}{2} : \frac{1}{2}

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

Cancel

\frac{2}{2} : \frac{1}{2}

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

\frac{1}{2} = \frac{2}{2}

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

\frac{1}{2} = \frac{2}{2}

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

- lo g_{10} (1.33 \times 1 0^{- 3})

lo g_{2} (\frac{x y}{z t})

2 ln (x) + 2 ln (x + 2)

lo g_{6} (30) + lo g_{6} (5)

lo g_{2} (5 ab)

What is 1/2 ln((1+(sqrt(2))/2)/(1-(sqrt(2))/2)) ?
The solution to 1/2 ln((1+(sqrt(2))/2)/(1-(sqrt(2))/2)) is (ln(1+(sqrt(2))/2)-ln(1-(sqrt(2))/2))/(2)