What is 0.9*log_{2}(0.9)+0.1*log_{2}(0.1) ?

The solution to 0.9*log_{2}(0.9)+0.1*log_{2}(0.1) is log_{2}(0.72246…)

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simplify 0.9log_{2}(0.9)+0.1log_{2}(0.1)

simplify

0.9 \cdot lo g_{2} (0.9) + 0.1 \cdot lo g_{2} (0.1)

lo g_{2} (0.72246 \dots)

Decimal Notation

- 0.46899 \dots

Solution steps

0.9 lo g_{2} (0.9) + 0.1 lo g_{2} (0.1)

Apply log rule:

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

0.9 lo g_{2} (0.9) = lo g_{2} (0. 9^{0.9})

= lo g_{2} (0. 9^{0.9}) + 0.1 lo g_{2} (0.1)

Apply log rule:

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

0.1 lo g_{2} (0.1) = lo g_{2} (0. 1^{0.1})

= lo g_{2} (0. 9^{0.9}) + lo g_{2} (0. 1^{0.1})

Apply log rule:

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

lo g_{2} (0. 9^{0.9}) + lo g_{2} (0. 1^{0.1}) = lo g_{2} (0. 9^{0.9} \cdot 0. 1^{0.1})

= lo g_{2} (0. 9^{0.9} \cdot 0. 1^{0.1})

0. 9^{0.9} \cdot 0. 1^{0.1} = 0.72246 \dots

= lo g_{2} (0.72246 \dots)

ln (2) - \frac{1}{2} - ln (\frac{1}{6}) - \frac{1}{24}

\frac{79}{3.9} ln (\frac{32}{3})

ln ((e^{- x^{3}}) (3 x^{3}))

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

lo g_{5} (7 \cdot 3)

What is 0.9*log_{2}(0.9)+0.1*log_{2}(0.1) ?
The solution to 0.9*log_{2}(0.9)+0.1*log_{2}(0.1) is log_{2}(0.72246…)