Question

Find the root that is a real number. rootindex 4 startroot 4096 endroot

Answer 1

the answer would be C.

Answer 2

Answer: 8

Step-by-step explanation:$\sqrt[4]{4096}$

We need to factor this, so

$4096|2\\2048|2\\1024|2\\512|2\\256|2\\128|2\\64|2\\32|2\\16|2\\8|2\\4|2\\2|2\\1$

$4096 = 2^{12}$

We have to remember that $x^{n+m}=x^{n}.x^{m}$

This way, we have $\sqrt[4]{4096}=\sqrt[4]{2^{12}} = \sqrt[4]{2^{4}.2^{4}.2^{4}}  = 2.2.2=8$