What is 2^{-n}(2^n-2^{1+n}) ?

The solution to 2^{-n}(2^n-2^{1+n}) is (2^n-2^{1+n})/(2^n)

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simplify 2^{-n}(2^n-2^{1+n})

simplify

2^{- n} (2^{n} - 2^{1 + n})

\frac{2 ^{n} - 2 ^{1 + n}}{2 ^{n}}

Solution steps

2^{- n} (2^{n} - 2^{1 + n})

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

2^{- n} = \frac{1}{2 ^{n}}

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

Apply the fraction rule:

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

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

Apply rule:

1 \cdot a = a

1 \cdot (2^{n} - 2^{1 + n}) = (2^{n} - 2^{1 + n})

= \frac{2 ^{n} - 2 ^{1 + n}}{2 ^{n}}

x^{2} - 10 = 4 x + 4

t^{2} - 10 t + 25 = 0

lo g_{10} (3)

4 a^{2} - 12 a + 9

\frac{8}{3}

What is 2^{-n}(2^n-2^{1+n}) ?
The solution to 2^{-n}(2^n-2^{1+n}) is (2^n-2^{1+n})/(2^n)