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Popular >

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

Solution

simplify

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

Solution

(n + 1)^{2}

Solution steps

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

Factor the number:

4 = 2^{2}

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

Apply exponent rule:

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

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

Apply exponent rule:

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

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

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

Apply log rule:

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

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

= (n + 1)^{2}

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Frequently Asked Questions (FAQ)

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