What is expand 5(2^x-x)^4(2^xln(2)-1) ?

The solution to expand 5(2^x-x)^4(2^xln(2)-1) is

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expand 5(2^x-x)^4(2^xln(2)-1)

expand

5 (2^{x} - x)^{4} (2^{x} ln (2) - 1)

5 ln (2) \cdot 3 2^{x} - 5 ln (2) \cdot 2^{4 x + 2} x + 15 ln (2) \cdot 2^{3 x + 1} x^{2} - 5 ln (2) \cdot 2^{2 x + 2} x^{3} + 5 ln (2) \cdot 2^{x} x^{4} - 5 \cdot 1 6^{x} + 5 \cdot 2^{3 x + 2} x - 15 \cdot 2^{2 x + 1} x^{2} + 5 \cdot 2^{2 + x} x^{3} - 5 x^{4}

Solution steps

5 (2^{x} - x)^{4} (2^{x} ln (2) - 1)

= 5 (2^{x} - x)^{4} (ln (2) \cdot 2^{x} - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 5 (2^{x} - x)^{4}, b = 2^{x} ln (2), c = 1

= 5 (2^{x} - x)^{4} \cdot 2^{x} ln (2) - 5 (2^{x} - x)^{4} \cdot 1

= 5 ln (2) \cdot 2^{x} (2^{x} - x)^{4} - 5 \cdot 1 \cdot (2^{x} - x)^{4}

Simplify

5 ln (2) \cdot 2^{x} (2^{x} - x)^{4} - 5 \cdot 1 \cdot (2^{x} - x)^{4} : 5 ln (2) \cdot 3 2^{x} - 5 ln (2) \cdot 2^{4 x + 2} x + 15 ln (2) \cdot 2^{3 x + 1} x^{2} - 5 ln (2) \cdot 2^{2 x + 2} x^{3} + 5 ln (2) \cdot 2^{x} x^{4} - 5 \cdot 1 6^{x} + 5 \cdot 2^{3 x + 2} x - 15 \cdot 2^{2 x + 1} x^{2} + 5 \cdot 2^{2 + x} x^{3} - 5 x^{4}

= 5 ln (2) \cdot 3 2^{x} - 5 ln (2) \cdot 2^{4 x + 2} x + 15 ln (2) \cdot 2^{3 x + 1} x^{2} - 5 ln (2) \cdot 2^{2 x + 2} x^{3} + 5 ln (2) \cdot 2^{x} x^{4} - 5 \cdot 1 6^{x} + 5 \cdot 2^{3 x + 2} x - 15 \cdot 2^{2 x + 1} x^{2} + 5 \cdot 2^{2 + x} x^{3} - 5 x^{4}

\frac{- a c x ^{2} + b}{( c x ^{2} + 16 ) ^{2}}

3 (sin (t) + t cos (t))

x^{''} + 100 \cdot x \cdot (1 + 10 x^{2})

\frac{v}{( x ( x + ( a + b ) ) ( x + ( a - b ) ) )}

(z^{2} + 2 z - 7 i)^{2}

What is expand 5(2^x-x)^4(2^xln(2)-1) ?
The solution to expand 5(2^x-x)^4(2^xln(2)-1) is