Question

Which function has a simplified base of 4^sqrt 4 ?

f(x) = 2^(3 sqrt 16)

f(x) = 2^(3 sqrt 64)

f(x) = 4(^3 sqrt 12 )^ 2x

f(x) = 4^(3 sqrt 64)^ 2x

Answer 1

c is your answer just took the quiz

Explanation:

Answer 2

f(x) =
`4^{(\sqrt[3]{64} )^(2x)}`.

Explanation:

We are given functions

`f(x) = 2^{(\sqrt[3]{16} )}`

f(x) =
`2^{(\sqrt[3]{64} )}`

f(x) =
`4{(\sqrt[3]{12})^(2x)}`

f(x) =
`4^{(\sqrt[3]{64} )^(2x)}`.

We need to find the function with simplified base of 4.

Let us simplify each of the function one by one.

On simplifying exponent in
`2^{(\sqrt[3]{16} )}`, we get
`2^{(2\sqrt[3]{2} )}`.

On simplifying exponent in
`2^{(\sqrt[3]{64} )}` we get 2^{4}.

On simplifying
`4{(\sqrt[3]{12})^(2x)}` we get
`4\left(\sqrt[3]{12}\:\:\right)\left(\sqrt[3]{12}\:\:\right)^x=4\left(\sqrt[3]{12\cdot 12}\right)\:^x=4\left(2\sqrt[3]{18}\right)^x`.

On simplifying
`4^{(\sqrt[3]{64} )^(2x)}` we get
`4^{\left(\sqrt[3]{64}\:\right)^(2x)}=4^{\left(4\:\right)^(2x)}=\:4^(16^x).`

Therefore, fourth function f(x) =
`4^{(\sqrt[3]{64} )^(2x)}` has simplified base 4.