What is 1/(2^x*ln(2))*1/2*(2^x)^2 ?

The solution to 1/(2^x*ln(2))*1/2*(2^x)^2 is (2^{x-1})/(ln(2))

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simplify 1/(2^xln(2))1/2*(2^x)^2

simplify

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

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

Solution steps

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

(2^{x})^{2} : 2^{x \cdot 2}

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

Multiply fractions:

a \cdot \frac{b}{c} \cdot \frac{d}{e} = \frac{a \cdot b \cdot d}{c \cdot e}

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

Multiply:

1 \cdot 1 \cdot 2^{x \cdot 2} = 2^{x \cdot 2}

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

2^{x} ln (2) \cdot 2 = ln (2) \cdot 2^{x + 1}

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

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = x^{a - b}

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

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

1.725 \cdot 1 0^{- 9} (4)

e \cdot e

e^{- t} \cdot cos (3 \cdot t)

x \frac{1 , A u}{196.97 g A u} x \frac{6.022 x ^{102}}{d x}

What is 1/(2^x*ln(2))*1/2*(2^x)^2 ?
The solution to 1/(2^x*ln(2))*1/2*(2^x)^2 is (2^{x-1})/(ln(2))