$\sqrt[5]{2^4\sqrt[3]{16}} = ? $

Question

$$\sqrt[5]{2^4\sqrt[3]{16}} = ? $$

Rewriting this and we have

$$\sqrt[5\cdot 3]{2^{4\cdot 3}4}$$

$$\sqrt[15]{2^{12}2^2}$$

Finally we get

$$\sqrt[15]{2^{12}2^2} = \sqrt[15]{2^{14}} = 2^{\frac{15}{14}}$$

Am I right?

Answer 1

\begin{align} \sqrt[5]{2^4 \sqrt[3]{16}} &=\sqrt[5]{2^4 \sqrt[3]{2^4}} =\sqrt[5]{2^4 \sqrt[3]{2^3 \cdot 2}} =\sqrt[5]{2^4 \cdot 2 \sqrt[3]{2}}\\ &=\sqrt[5]{2^5 \sqrt[3]{2}} =2 \sqrt[5]{\sqrt[3]{2}} =2 \cdot \sqrt[15]{2} =2 \cdot 2^{1/15}\\ &=2^{16/15}. \end{align}

Answer 2

$$2^4\sqrt[3]{16}=2^4\cdot2^{4/3}=2^{16/3}\implies\sqrt[5]{2^4\sqrt{16}}=\left(2^{16/3}\right)^{1/5}=2^{16/15}$$

so no, you are not right...but almost.

Answer 3

In your first reformulation, you should get $\sqrt[15]{2^{4 \cdot 3} \cdot 16}$ and in the last part note that $\sqrt[a]{c^b} = c^{\frac{b}{a}}$ and not $c^{\frac{a}{b}}$.

Answer 4

$$\sqrt[5]{2^4\sqrt[3]{16}} = (2^4)^{1/5}(16^{1/3})^{1/5}=2^{4/5}(2^4)^{1/15}=2^{4/5}2^{4/15}=2^{16/15}$$

Answer 5

You wrote $$\sqrt[5]{2^4\sqrt[3]{16}}=\sqrt[5\cdot 3]{2^{4\cdot 3}4}$$ which is not true because
$$\sqrt[3]{16}=16^\frac{1}{3}$$
That there is the third root of $16$, not $2^{4\cdot 3}$. Also you can not write $$\sqrt[5]{something\cdot \sqrt[3]{something}}=\sqrt[5\cdot 3]{something\cdot something}$$
The way to do it is:
$$\sqrt[5]{2^4\sqrt[3]{16}}=\sqrt[5]{2^4\cdot16^\frac{1}{3}}=\sqrt[5]{2^4\cdot2^\frac{4}{3}}=\sqrt[5]{2^\frac{16}{3}}=2^\frac{\frac{16}{3}}{5}=2^\frac{16}{15}$$

Answer 6

$\sqrt[5]{2^4\sqrt[3]{16}} =$

$\sqrt[5\cdot 3]{2^{4\cdot 3}\color{blue}{16}}=$

$\sqrt[15]{2^{12}2^{\color{blue}{4}}}=$

$\sqrt[15]{2^{12}2^{\color{blue}{4}}} = \sqrt[15]{2^{\color{blue}{16}}} = 2^{\frac{\not 1\not 5\color{red}{16}}{\not{\color{blue}{1\not 6}}\color{red}{15}}}$

But in my opinion your method seems a little scatter-shot and undirected. In particular $\sqrt[k]{b} = \sqrt[mk]{b^m}$ rubs me the wrong way. It's not wrong per se, but is seems that we are going in the wrong direction and making things complicated rather than simpler. And I fear extraneous roots and sign errors. (Ex: $\sqrt[5]{(-1)^3} \ne \sqrt [5*2]{(-1)^{3*2}}$).

Although there is no universal right or wrong way to do things try to develop a more systematic approach. I'd personal reduce to a common base, convert radicals to fractional exponents, and then just do the math:

$\sqrt[5]{2^4\sqrt[3]{16}}=$

$\sqrt[5]{2^4\sqrt[3]{2^4}} =$

$(2^4(2^4)^{\frac 13})^{\frac 15}=$

$2^{\frac 15(4 + 4*\frac 13)} =$

$2^{\frac{16}{15}}$.

Also, I suppose I should point out that:

$2^{\frac {16}{15}} = 2^{1 \frac 1{15}} = 2\sqrt[15]{2}$

which could be an acceptable answer. As is $2\times 2^{\frac 1{15}}$.

Which of these three answers $2^{\frac {16}{15}}, 2\sqrt[15]{2}, 2\times 2^{\frac 1{15}}$ it the correct one? Well, none are, or they all are.

I personally prefer $2\sqrt[15]{2}$ as it.... well, I get a better sense of what makes the number. I would prefer $2\times 2^{\frac 1{15}}$ as I'd general like to conform radicals and exponents, but in this case requiring the $\times$ sign bugs me.

But this is purely subjective.