Rules of Roots Combined: Number of terms

Solve the following exercise:

$\sqrt{2}\cdot\sqrt{5}\cdot\sqrt{2}\cdot\sqrt{2}=$

In order to simplify the given expression we use two laws of exponents:

A. Defining the root as an exponent:

$\sqrt[n]{a}=a^{\frac{1}{n}}$B. The law of exponents for a product of numbers with the same base (in the opposite direction):

$x^n\cdot y^n =(x\cdot y)^n$

Let's start by definging the roots as exponents using the law of exponents shown in A:

$\sqrt{2}\cdot\sqrt{5}\cdot\sqrt{2}\cdot\sqrt{2}= \\ \downarrow\\ 2^{\frac{1}{2}}\cdot5^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot2^{\frac{1}{2}}=$Since we are multiplying between four numbers with the same exponents we can use the law of exponents shown in B (which also applies to a product of numbers with the same base) and combine them together in a product wit the same base which is raised to the same exponent:

$2^{\frac{1}{2}}\cdot5^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot2^{\frac{1}{2}}= \\ (2\cdot5\cdot2\cdot2)^{\frac{1}{2}}=\\ 40^{\frac{1}{2}}=\\ \boxed{\sqrt{40}}$In the last step we performed the product which is in the base, then we used again the definition of the root as an exponent shown earlier in A (in the opposite direction) to return to writing the root.

Therefore, note that the correct answer is answer C.

$\sqrt{40}$

Solve the following exercise:

$\sqrt{4}\cdot\sqrt{2}\cdot\sqrt{2}=$

4

Complete the following exercise:

$\sqrt[3]{\sqrt{16}}\cdot\sqrt[]{\sqrt{16}}=$

$16^{\frac{5}{12}}$

Complete the following exercise:

$\sqrt{\sqrt{4}}\cdot\sqrt{\sqrt{2}}=$

$2^{\frac{1}{4}}\sqrt{2}$

Complete the following exercise:

$\sqrt{\sqrt{49}}\cdot\sqrt{\sqrt{16}}=$

$2\sqrt{7}$

Complete the following exercise:

$\sqrt{25}\cdot\sqrt[3]{\sqrt{25}}=$

$5^{1\frac{1}{3}}$

Solve the following exercise:

$\sqrt{1}\cdot\sqrt{2}\cdot\sqrt{3}=$

$\sqrt{6}$

Solve the following exercise:

$\sqrt{\frac{2}{4}}\cdot\sqrt{\frac{8}{16}}=$

$\frac{1}{2}$

Complete the following exercise:

$\sqrt{\sqrt{16}}\cdot\sqrt{\sqrt{8}}=$

$$$\sqrt[4]{128}$

Complete the following exercise:

$\sqrt{\sqrt{2}}\cdot\sqrt{\sqrt{4}}=$

$\sqrt[4]{8}$

Solve the following exercise:

$\sqrt{2}\cdot\sqrt{2}\cdot\sqrt{0}=$

$0$

Solve the following exercise:

$\frac{\sqrt{12}\cdot\sqrt{4}\cdot\sqrt{3}}{\sqrt{2}\cdot\sqrt{2}}=$

6

Solve the following exercise:

$\sqrt{10}\cdot\sqrt{2}\cdot\sqrt{5}=$

$10$

Solve the following exercise:

$\frac{\sqrt{81}\cdot\sqrt{4}}{\sqrt{9}\cdot\sqrt{9}}=$

$2$

Complete the following exercise:

$\sqrt[3]{\sqrt{25}}\cdot\sqrt[3]{\sqrt{64}}=$

$2\sqrt[3]{5}$

Complete the following exercise:

$\sqrt[5]{\sqrt{3}}\cdot\sqrt[6]{\sqrt{3}}=$

$3^{\frac{1}{10}+\frac{1}{12}}$

Solve the following exercise:

$\sqrt{\frac{2}{4}}\cdot\sqrt{6}=$

$\sqrt{3}$

Solve the following exercise:

$\frac{\sqrt{10}\cdot\sqrt{5}\cdot\sqrt{2}}{\sqrt{5}\cdot\sqrt{5}\cdot\sqrt{4}}=$

$1$

Complete the following exercise:

$\sqrt[3]{\sqrt{3}}\cdot\sqrt[3]{\sqrt{4}}=$

$\sqrt[6]{12}$

Solve the following exercise:

$\sqrt{5}\cdot\sqrt{10}\cdot\sqrt{2}\cdot\sqrt{4}=$

$20$