What is ((2^{n+4}*4*2^{n-1}))/((2^n)) ?

The solution to ((2^{n+4}*4*2^{n-1}))/((2^n)) is 2^{n+5}

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simplify ((2^{n+4}42^{n-1}))/((2^n))

simplify

\frac{( 2 ^{n + 4} \cdot 4 \cdot 2 ^{n - 1} )}{( 2 ^{n} )}

2^{n + 5}

Solution steps

\frac{2 ^{n + 4} \cdot 4 \cdot 2 ^{n - 1}}{2 ^{n}}

Simplify

\frac{2 ^{n + 4} \cdot 2 ^{n - 1}}{2 ^{n}} : 2^{n + 3}

= 2^{n + 3} \cdot 4

Simplify

2^{n + 3} \cdot 4 : 2^{n + 5}

= 2^{n + 5}

3 x^{4} + 6 x^{3} - 2 x^{3} - 10 x - 5

13 p^{5} \cdot q^{2} - 13 p^{2} \cdot q^{2}

(- a)^{\frac{3}{2}}

8 x^{3} - 6 x^{2} + 4 x

5 x^{- 6} - 9 x^{2} - 7

What is ((2^{n+4}*4*2^{n-1}))/((2^n)) ?
The solution to ((2^{n+4}*4*2^{n-1}))/((2^n)) is 2^{n+5}