What is simplify-0.05log_{2}(0.05)-0.95log_{2}(0.95) ?

The solution to simplify-0.05log_{2}(0.05)-0.95log_{2}(0.95) is -log_{2}(0.81994…)

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simplify-0.05log_{2}(0.05)-0.95log_{2}(0.95)

simplify

- 0.05 lo g_{2} (0.05) - 0.95 lo g_{2} (0.95)

- lo g_{2} (0.81994 \dots)

Decimal Notation

0.28639 \dots

Solution steps

- 0.05 lo g_{2} (0.05) - 0.95 lo g_{2} (0.95)

Apply log rule:

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

0.05 lo g_{2} (0.05) = lo g_{2} (0.0 5^{0.05})

= - lo g_{2} (0.0 5^{0.05}) - 0.95 lo g_{2} (0.95)

Apply log rule:

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

0.95 lo g_{2} (0.95) = lo g_{2} (0.9 5^{0.95})

= - lo g_{2} (0.0 5^{0.05}) - lo g_{2} (0.9 5^{0.95})

Rewrite

- lo g_{2} (0.0 5^{0.05}) - lo g_{2} (0.9 5^{0.95})

- (lo g_{2} (0.0 5^{0.05}) + lo g_{2} (0.9 5^{0.95}))

= - (lo g_{2} (0.0 5^{0.05}) + lo g_{2} (0.9 5^{0.95}))

Apply log rule:

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

lo g_{2} (0.0 5^{0.05}) + lo g_{2} (0.9 5^{0.95}) = lo g_{2} (0.0 5^{0.05} \cdot 0.9 5^{0.95})

= - lo g_{2} (0.0 5^{0.05} \cdot 0.9 5^{0.95})

0.0 5^{0.05} \cdot 0.9 5^{0.95} = 0.81994 \dots

= - lo g_{2} (0.81994 \dots)

ln (\frac{x x + 1}{2 e ^{4}})

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lo g_{3} (\frac{( x + 4 ) ^{2}}{x ^{2} + y ^{2}})

2 lo g_{6} (u) + 7 lo g_{6} (v)

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What is simplify-0.05log_{2}(0.05)-0.95log_{2}(0.95) ?
The solution to simplify-0.05log_{2}(0.05)-0.95log_{2}(0.95) is -log_{2}(0.81994…)