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

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

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simplify ((22^n-22^{n-2}))/((2^n-2^{n-1))}

simplify

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

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

Solution steps

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

2 \cdot 2^{n} = 2^{1 + n}

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

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

lo g_{2} (g) (\frac{1}{8})

- 20 z - 6 z - 15

(5 x^{4} - 6 x^{2}) \cdot x^{2}

- (3 a + 0 \cdot 1)^{2}

a^{2} - b^{2} - 10 a + 25

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