What is expand (1-p)log_{2}(1-p) ?

The solution to expand (1-p)log_{2}(1-p) is log_{2}(1-p)-plog_{2}(1-p)

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expand (1-p)log_{2}(1-p)

expand

(1 - p) lo g_{2} (1 - p)

lo g_{2} (1 - p) - p lo g_{2} (1 - p)

Solution steps

(1 - p) lo g_{2} (1 - p)

= lo g_{2} (1 - p) (1 - p)

Apply the distributive law:

a (b - c) = ab - a c

lo g_{2} (1 - p) (1 - p) = lo g_{2} (1 - p) \cdot 1 - lo g_{2} (1 - p) p

= lo g_{2} (1 - p) \cdot 1 - lo g_{2} (1 - p) p

Simplify

lo g_{2} (1 - p) \cdot 1 - lo g_{2} (1 - p) p : lo g_{2} (1 - p) - p lo g_{2} (1 - p)

= lo g_{2} (1 - p) - p lo g_{2} (1 - p)

(- 1, 4, - 2, 1) l e f t (3, 5, 2, - 1)

5 (- 7 + 14)

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

\frac{184 ( s + 5 )}{s ( s + 4 ) ( s ^{2} + 4 )}

\frac{\partial}{\partial x} ((z) (2 x^{3} - 4 x^{2} - 6 y))

What is expand (1-p)log_{2}(1-p) ?
The solution to expand (1-p)log_{2}(1-p) is log_{2}(1-p)-plog_{2}(1-p)