What is expand y(-1/(ln(2)2^y)=-1/(ln(2)2^x)) ?

The solution to expand y(-1/(ln(2)2^y)=-1/(ln(2)2^x)) is -y/(ln(2)*2^y)=-y/(ln(2)*2^x)

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expand y(-1/(ln(2)2^y)=-1/(ln(2)2^x))

expand

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

- \frac{y}{ln ( 2 ) \cdot 2 ^{y}} = - \frac{y}{ln ( 2 ) \cdot 2 ^{x}}

Solution steps

y (- \frac{1}{ln ( 2 ) \cdot 2 ^{y}} = - \frac{1}{ln ( 2 ) \cdot 2 ^{x}})

= y (- = \frac{1}{ln ( 2 ) \cdot 2 ^{y}} - \frac{1}{ln ( 2 ) \cdot 2 ^{x}})

Apply the distributive law:

a (b - c) = ab - a c

a = y, b = - \frac{1}{ln ( 2 ) \cdot 2 ^{y}} =, c = \frac{1}{ln ( 2 ) \cdot 2 ^{x}}

= y (- \frac{1}{ln ( 2 ) \cdot 2 ^{y}} =) - y \frac{1}{ln ( 2 ) \cdot 2 ^{x}}

Apply minus-plus rules

+ (- a) = - a

- \frac{1}{ln ( 2 ) \cdot 2 ^{y}} y = - \frac{1}{ln ( 2 ) \cdot 2 ^{x}} y

Simplify

- \frac{1}{ln ( 2 ) \cdot 2 ^{y}} y = - \frac{1}{ln ( 2 ) \cdot 2 ^{x}} y : - \frac{y}{ln ( 2 ) \cdot 2 ^{y}} = - \frac{y}{ln ( 2 ) \cdot 2 ^{x}}

- \frac{y}{ln ( 2 ) \cdot 2 ^{y}} = - \frac{y}{ln ( 2 ) \cdot 2 ^{x}}

\frac{20 ( s + 1 )}{s ( s + 5 ) ( s ^{2} + 2 s + 10 )}

\frac{d}{d t} ((P (t, a, b)) (a P (t, a, b) (b - P (t, a, b))))

d (6)

x H (x - 3)

5 ln (2 x + 3) + \frac{2 ( 5 x - 5 )}{2 x + 3}

What is expand y(-1/(ln(2)2^y)=-1/(ln(2)2^x)) ?
The solution to expand y(-1/(ln(2)2^y)=-1/(ln(2)2^x)) is -y/(ln(2)*2^y)=-y/(ln(2)*2^x)