What is 1/2 (-log_{2}(3)+1) ?

The solution to 1/2 (-log_{2}(3)+1) is -1/2 log_{2}(3)+1/2

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simplify 1/2 (-log_{2}(3)+1)

simplify

\frac{1}{2} (- lo g_{2} (3) + 1)

- \frac{1}{2} lo g_{2} (3) + \frac{1}{2}

Decimal Notation

- 0.29248 \dots

Solution steps

\frac{1}{2} (- lo g_{2} (3) + 1)

Apply the distributive law:

a (b + c) = ab + a c

\frac{1}{2} (- lo g_{2} (3) + 1) = \frac{1}{2} (- lo g_{2} (3)) + \frac{1}{2} \cdot 1

= \frac{1}{2} (- lo g_{2} (3)) + \frac{1}{2} \cdot 1

\frac{1}{2} (- lo g_{2} (3)) + \frac{1}{2} \cdot 1 = - \frac{1}{2} lo g_{2} (3) + \frac{1}{2}

= - \frac{1}{2} lo g_{2} (3) + \frac{1}{2}

- \frac{e ^{n θ} ( - n e ^{- n θ} θ - e ^{- n θ} )}{n}

tan (\frac{π}{3} - x)

(x + 5) (2 x - 35)

lo g_{12} (\frac{y ^{4}}{z m ^{4}})

\frac{( ( x ^{3} - 2 x ) ^{2} - 1 )}{x ^{2} - 1}

What is 1/2 (-log_{2}(3)+1) ?
The solution to 1/2 (-log_{2}(3)+1) is -1/2 log_{2}(3)+1/2