簡約 4^{log_{2}(n)+1}

解

簡約

4^{l o g_{2} (n) + 1}

4 n^{2}

解答ステップ

4^{l o g_{2} (n) + 1}

指数の規則を適用する:

a^{b + c} = a^{b} \cdot a^{c}

4^{l o g_{2} (n) + 1} = 4^{l o g_{2} (n)} \cdot 4

= 4^{l o g_{2} (n)} \cdot 4

数を因数に分解する:

4 = 2^{2}

= (2^{2})^{l o g_{2} (n)} \cdot 4

指数の規則を適用する:

(a^{b})^{c} = a^{b c}, a \geq 0

(2^{2})^{l o g_{2} (n)} = 2^{2 l o g_{2} (n)}

= 2^{2 l o g_{2} (n)} \cdot 4

指数の規則を適用する:

a^{b c} = (a^{b})^{c}, a \geq 0

2^{2 l o g_{2} (n)} = (2^{l o g_{2} (n)})^{2}

= (2^{l o g_{2} (n)})^{2} \cdot 4

対数の規則を適用する:

a^{l o g_{a} (b)} = b

2^{l o g_{2} (n)} = n

= n^{2} \cdot 4

= 4 n^{2}