Taking factors out of a radical

Text

Take out of the radical sign every power you can:

• (a) $\sqrt[5]{2^{12}}$
• (b) $\sqrt[6]{3^{24}}$
• (c) $\sqrt[3]{2^7\cdot 3^8}$
• (d) $\sqrt[7]{2^{11}\cdot 5^{45}\cdot 7^{32}}$

Solution

From $\sqrt[n]{a^{m}}=a^{\lfloor m/n\rfloor}\sqrt[n]{a^{\,m\bmod n}}$: extract the quotient of the exponent divided by the index; the remainder stays under the root.

(a) $12=5\cdot 2+2$: $\sqrt[5]{2^{12}}=2^2\sqrt[5]{2^2}=4\sqrt[5]{4}$.

(b) $24=6\cdot 4$ exactly: $\sqrt[6]{3^{24}}=3^4=81$ (exact root, no remainder).

(c) $7=3\cdot 2+1$, $8=3\cdot 2+2$: $\sqrt[3]{2^7\cdot 3^8}=2^2\cdot 3^2\sqrt[3]{2\cdot 3^2}=36\sqrt[3]{18}$.

(d) $11=7\cdot 1+4$, $45=7\cdot 6+3$, $32=7\cdot 4+4$: $\sqrt[7]{2^{11}\cdot 5^{45}\cdot 7^{32}}=2\cdot 5^{6}\cdot 7^{4}\sqrt[7]{2^{4}\cdot 5^{3}\cdot 7^{4}}$.

$\boxed{\;(a)\;4\sqrt[5]{4}\qquad (b)\;81\qquad (c)\;36\sqrt[3]{18}\qquad (d)\;2\cdot 5^{6}\cdot 7^{4}\sqrt[7]{2^{4}\cdot 5^{3}\cdot 7^{4}}\;}$