What is 4^{log_{2}(n)+1} ?

The solution to 4^{log_{2}(n)+1} is 4n^2

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simplify 4^{log_{2}(n)+1}

simplify

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

4 n^{2}

Solution steps

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

Apply exponent rule:

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

Factor the number:

4 = 2^{2}

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

Apply exponent rule:

(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

Apply exponent rule:

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

Apply log rule:

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

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

= n^{2} \cdot 4

= 4 n^{2}

\frac{3}{2 3}

a^{2} a^{3}

5^{2.5}

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

4.98 \cdot 1 0^{15} \cdot e^{0.47 (2000 - 2010)}