Rules of Roots Combined: Number of terms

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

a. The definition of root as an exponent:

$\sqrt[n]{a}=a^{\frac{1}{n}}$

b. The law of exponents for exponents applied to terms in parentheses (in reverse order):

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

Let's start by converting the square roots to exponents using the law of exponents mentioned in a':

$\sqrt{1}\cdot\sqrt{2}\cdot\sqrt{3} \\ \downarrow\\ 1^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot3^{\frac{1}{2}}=$

We'll continue, since there is multiplication between three terms with identical exponents, we can use the law of exponents mentioned in b' (which also applies to multiplication of several terms in parentheses) and combine them together in multiplication under parentheses raised to the same exponent:

$1^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot3^{\frac{1}{2}}= \\ (1\cdot2\cdot3)^{\frac{1}{2}}=\\ 6^{\frac{1}{2}}=\\ \boxed{\sqrt{6}}$

In the final steps, we performed the multiplication within the parentheses and again used the definition of root as an exponent mentioned in a' (in reverse order) to return to root notation.

Therefore, the correct answer is answer d.

Notice that in the given problem, a multiplication is performed between three terms, one of which is:

$\sqrt{0}$and let's remember that the root (of any order) of the number 0 is 0, meaning that:

$\sqrt{0}=0$and since multiplying any number by 0 will always yield the result 0,

therefore the result of the multiplication in the problem is 0, meaning:

$\sqrt{2}\cdot\sqrt{2}\cdot\sqrt{0}=\\ \sqrt{2}\cdot\sqrt{2}\cdot0=\\ \boxed{0}$and thus the correct answer is answer C.

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

a. The definition of root as an exponent:

$\sqrt[n]{a}=a^{\frac{1}{n}}$

b. The law of exponents for an exponent applied to a product in parentheses (in reverse direction):

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

Let's start by converting the square roots to exponents using the law of exponents mentioned in a':

$\sqrt{4}\cdot\sqrt{2}\cdot\sqrt{2} \\ \downarrow\\ 4^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot2^{\frac{1}{2}}=$

We'll continue, since we have a multiplication of three terms with identical exponents, we can use the law of exponents mentioned in b' (which also applies to multiplying several terms in parentheses) and combine them together in a multiplication under parentheses that are raised to the same exponent:

$4^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot2^{\frac{1}{2}}= \\ (4\cdot2\cdot2)^{\frac{1}{2}}=\\ 16^{\frac{1}{2}}=\\ \sqrt{16}=\\ \boxed{4}$

In the final steps, we first performed the multiplication within the parentheses, then we used again the definition of root as an exponent mentioned in a' (in reverse direction) to return to root notation, and in the final stage, we calculated the known square root of 16.

Therefore, we can identify that the correct answer is answer c.

To find the value of $\sqrt{\sqrt{49}} \cdot \sqrt{\sqrt{16}}$, we will follow these steps:

• Step 1: Simplify $\sqrt{\sqrt{49}}$.
• Step 2: Simplify $\sqrt{\sqrt{16}}$.
• Step 3: Multiply the simplified results together.

Step 1: $\sqrt{\sqrt{49}}$
- Calculate $\sqrt{49} = 7$.
- Therefore, $\sqrt{\sqrt{49}} = \sqrt{7}$.

Step 2: $\sqrt{\sqrt{16}}$
- Calculate $\sqrt{16} = 4$.
- Therefore, $\sqrt{\sqrt{16}} = \sqrt{4} = 2$.

Step 3: Multiply the simplified results:
- Multiply $\sqrt{7}$ by $2$.
- The product is $2 \cdot \sqrt{7} = 2\sqrt{7}$.

Therefore, the value of $\sqrt{\sqrt{49}} \cdot \sqrt{\sqrt{16}}$ is $\mathbf{2\sqrt{7}}$.

To solve this problem, we'll follow these steps:

• Step 1: Simplify each term individually.
• Step 2: Multiply the simplified terms together.
• Step 3: Compare with choices if necessary.

Let's begin:

Step 1: Simplify each term:

The expression is $\sqrt{\sqrt{2}} \cdot \sqrt{\sqrt{4}}$.

- Simplifying $\sqrt{\sqrt{2}}$: A root of a root involves multiplying the indices. We have $\sqrt[2]{\sqrt[2]{2}}$, which becomes $\sqrt[4]{2}$.

- Simplifying $\sqrt{\sqrt{4}}$: Note that $\sqrt{4} = 2$, so $\sqrt{2} = \sqrt{2}$.

Conclusively, $\sqrt{\sqrt{4}} = \sqrt{2}$.

Step 2: Multiply the simplified terms:

Now, multiply $\sqrt[4]{2} \times \sqrt{2}$:

$\sqrt[4]{2} \cdot \sqrt{2} = \sqrt[4]{2 \times 2^{1/2}} = \sqrt[4]{2^{3/2}} = \sqrt[4]{8}$.

Therefore, our simplified expression is $\sqrt[4]{8}$.

Step 3: Compare with answer choices:

The correct choice is $\sqrt[4]{8}$, matching choice 3.

Therefore, the solution to the problem is $\sqrt[4]{8}$.

To solve the problem $\sqrt{\sqrt{16}}\cdot\sqrt{\sqrt{8}}$, we will follow these steps:

• Step 1: Simplify each nested square root expression.
• Step 2: Multiply the simplified expressions of the roots.

Step 1: Evaluate $\sqrt{\sqrt{16}}$.
Since 16 can be expressed as $2^4$, we have: $\sqrt{\sqrt{16}} = \left(16^{1/2}\right)^{1/2} = 16^{1/4} = \left(2^4\right)^{1/4} = 2^{4/4} = 2^{1} = 2$

Evaluate $\sqrt{\sqrt{8}}$.
Since 8 can be expressed as $2^3$, we have: $\sqrt{\sqrt{8}} = \left(8^{1/2}\right)^{1/2} = 8^{1/4} = \left(2^3\right)^{1/4} = 2^{3/4}$

Step 2: Multiply these simplified expressions together: $2 \cdot 2^{3/4} = 2^{1} \cdot 2^{3/4} = 2^{1 + 3/4} = 2^{7/4}$

Finally, converting back to radical form: $2^{7/4} = \left(2^7\right)^{1/4} = \sqrt[4]{2^7} = \sqrt[4]{128}$

Thus, the solution to the problem is $\sqrt[4]{128}$, which corresponds to answer choice 3.

To solve the given exercise, let's simplify each square root expression separately:

• Step 1: Simplify $\sqrt{\frac{2}{4}}$.

  The fraction $\frac{2}{4}$ simplifies to $\frac{1}{2}$. Thus, $\sqrt{\frac{2}{4}} = \sqrt{\frac{1}{2}} = \frac{\sqrt{1}}{\sqrt{2}} = \frac{1}{\sqrt{2}}$.

• Step 2: Simplify $\sqrt{\frac{8}{16}}$.

  The fraction $\frac{8}{16}$ simplifies to $\frac{1}{2}$. Thus, $\sqrt{\frac{8}{16}} = \sqrt{\frac{1}{2}} = \frac{\sqrt{1}}{\sqrt{2}} = \frac{1}{\sqrt{2}}$.

• Step 3: Multiply the results from Step 1 and Step 2.

  $\sqrt{\frac{2}{4}} \cdot \sqrt{\frac{8}{16}} = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{1 \cdot 1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$.

Therefore, the solution to the given expression is $\frac{1}{2}$.

To solve the expression $\sqrt{\sqrt{4}}\cdot\sqrt{\sqrt{2}}$, we will use properties of exponents and roots.

First, let's simplify each part:

• $\sqrt{\sqrt{4}}$:

We know $\sqrt{4} = 2$. Therefore, $\sqrt{\sqrt{4}}$ can be rewritten as $\sqrt{2}$, because $\sqrt{4} = 2$ and further taking square root gives $2^{1/2}$.

• $\sqrt{\sqrt{2}}$:

This expression is equivalent to $(2^{1/2})^{1/2}$. Using the property $(a^{m})^{n} = a^{m \cdot n}$, we have:

$(2^{1/2})^{1/2} = 2^{1/4}$.

Now, the original expression simplifies to:

$\sqrt{2} \cdot 2^{1/4}$

This product is expressed as:

$2^{1/2} \cdot 2^{1/4}$. When multiplying like bases, add the exponents:

$2^{1/2 + 1/4} = 2^{2/4 + 1/4} = 2^{3/4}$

Thus, the final expression is:

$2^{1/4}\sqrt{2}$.

Comparing this to the choices provided, the correct answer is:

$2^{1/4}\sqrt{2}$ (Choice 3).

Therefore, the solution to the problem is $\boxed{2^{\frac{1}{4}}\sqrt{2}}$.

To solve the problem $\sqrt{25} \cdot \sqrt[3]{\sqrt{25}}$, follow these steps:

• First, express each root in terms of exponents:
• $\sqrt{25} = 25^{1/2}$
• $\sqrt[3]{\sqrt{25}} = \sqrt[3]{25^{1/2}}$
• Using the law of exponents $(a^m)^n = a^{m \cdot n}$, simplify $\sqrt[3]{25^{1/2}}$ as follows:
• $(25^{1/2})^{1/3} = 25^{(1/2) \cdot (1/3)} = 25^{1/6}$
• Now, multiply the two expressions:
• $25^{1/2} \cdot 25^{1/6} = 25^{(1/2 + 1/6)}$
• Calculate the sum of the exponents: $\frac{1}{2} + \frac{1}{6} = \frac{3}{6} + \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$
• This gives: $25^{2/3}$
• Recognize $25 = 5^2$, thus: $(5^2)^{2/3} = 5^{2 \cdot (2/3)} = 5^{4/3}$
• Convert mixed fraction: $5^{4/3} = 5^{1 + 1/3} = 5^{1\frac{1}{3}}$

Therefore, the product $\sqrt{25} \cdot \sqrt[3]{\sqrt{25}}$ equals $\mathbf{5^{1\frac{1}{3}}}$.

To solve this problem, we will simplify the expression $\sqrt[3]{\sqrt{16}} \cdot \sqrt[]{\sqrt{16}}$ using the rules for exponents and roots.

First, consider the inner square root $\sqrt{16}$. We know that:
$\sqrt{16} = 16^{1/2}$

Next, we address the cube root term $\sqrt[3]{\sqrt{16}}$. Express $\sqrt{16}$ as $16^{1/2}$, then:

• $\sqrt[3]{\sqrt{16}} = \sqrt[3]{16^{1/2}}$ Convert to exponents: $\sqrt[3]{16^{1/2}} = (16^{1/2})^{1/3} = 16^{(1/2) \cdot (1/3)} = 16^{1/6}$
• The other term is $\sqrt{\sqrt{16}} = \sqrt{16^{1/2}}$ Convert to exponents: $\sqrt{16^{1/2}} = (16^{1/2})^{1/2} = 16^{(1/2) \cdot (1/2)} = 16^{1/4}$

Now, multiply these results:
$16^{1/6} \cdot 16^{1/4}$

Using the product rule for exponents $(a^m \cdot a^n = a^{m+n})$, combine the exponents:
$16^{1/6 + 1/4}$

Find the common denominator to add the fractions:

• $1/6 = 2/12$
• $1/4 = 3/12$ $1/6 + 1/4 = 2/12 + 3/12 = 5/12$

Thus, the expression becomes:
$16^{5/12}$

Therefore, the simplified expression is $16^{\frac{5}{12}}$.

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.

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

a. The definition of root as an exponent:

$\sqrt[n]{a}=a^{\frac{1}{n}}$

b. The law of exponents for an exponent applied to a product in parentheses (in reverse direction):

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

Let's start by converting the square roots to exponents using the law of exponents mentioned in a':

$\sqrt{2}\cdot\sqrt{2}\cdot\sqrt{2}\cdot\sqrt{1}\cdot\sqrt{1}= \\ \downarrow\\ 2^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot1^{\frac{1}{2}}\cdot1^{\frac{1}{2}}=$

We'll continue, since there is a multiplication between five terms with identical exponents we can use the law of exponents mentioned in b' (which of course also applies to multiplying several terms in parentheses) and combine them together in a multiplication under parentheses which are raised to the same exponent:

$2^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot1^{\frac{1}{2}}\cdot1^{\frac{1}{2}}= \\ (2\cdot2\cdot2\cdot1\cdot1)^{\frac{1}{2}}=\\ 8^{\frac{1}{2}}=\\ \boxed{\sqrt{8}}$

In the final steps, we first performed the multiplication within the parentheses, then we used again the definition of root as an exponent mentioned earlier in a' (in reverse direction) to return to root notation.

Therefore, we can identify that the correct answer is answer a'.

To solve the problem, we'll simplify the expression $\frac{\sqrt{81}\cdot\sqrt{4}}{\sqrt{9}\cdot\sqrt{9}}$ using square root properties and arithmetic operations.

Step 1: Simplify each square root individually:

• $\sqrt{81} = 9$, $\sqrt{4} = 2$.
• Both $\sqrt{9}$ terms are equal to 3.

Step 2: Perform multiplication of numbers:

• Numerator: $9 \times 2 = 18$.
• Denominator: $3 \times 3 = 9$.

Step 3: Simplify the fraction:

• $\frac{18}{9} = 2$, after dividing both the numerator and the denominator by the GCD, which is 9.

Therefore, the solution to the problem is $2$.

To solve the expression $\sqrt{\frac{2}{4}} \cdot \sqrt{6}$, we will break it down and simplify step by step.

Step 1: Simplify the square root of the fraction.
$\sqrt{\frac{2}{4}}$ can be rewritten using the square root of a quotient property:
$\sqrt{\frac{2}{4}} = \frac{\sqrt{2}}{\sqrt{4}}$.

Step 2: Simplify $\sqrt{4}$.
Since $\sqrt{4} = 2$, the expression becomes:
$\frac{\sqrt{2}}{2}$.

Step 3: Multiply by $\sqrt{6}$.
Now multiply $\frac{\sqrt{2}}{2}$ by $\sqrt{6}$:
$\frac{\sqrt{2}}{2} \cdot \sqrt{6} = \frac{\sqrt{2 \cdot 6}}{2}$.

Step 4: Simplify the square root.
The multiplication inside the square root becomes $\sqrt{12}$, so:
$\frac{\sqrt{12}}{2}$.

Step 5: Simplify $\sqrt{12}$.
Since $\sqrt{12} = \sqrt{4 \cdot 3} = \sqrt{4} \cdot \sqrt{3} = 2\sqrt{3}$,
this results in $\frac{2\sqrt{3}}{2} = \sqrt{3}$.

Therefore, the solution to the problem is $\sqrt{3}$.

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

a. The definition of root as an exponent:

$\sqrt[n]{a}=a^{\frac{1}{n}}$

b. The law of exponents for an exponent applied to a product in parentheses (in reverse direction):

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

Let's start with converting the square roots to exponents using the law of exponents mentioned in a:

$\sqrt{10}\cdot\sqrt{2}\cdot\sqrt{5} \\ \downarrow\\ 10^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot5^{\frac{1}{2}}=$

We'll continue, since there is a multiplication between three terms with identical exponents we can use the law of exponents mentioned in b (which also applies to multiplying several terms in parentheses) and combine them together in a multiplication under parentheses which are raised to the same exponent:

$10^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot5^{\frac{1}{2}}= \\ (10\cdot2\cdot5)^{\frac{1}{2}}=\\ 100^{\frac{1}{2}}=\\ \sqrt{100}=\\ \boxed{10}$

In the final steps, we first performed the multiplication within the parentheses, then we used again the definition of root as an exponent mentioned in a (in reverse direction) to return to root notation, and in the final stage we calculated the known square root of 100.

Therefore, we can identify that the correct answer (most appropriate) is answer d.

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

a. Root definition as an exponent:

$\sqrt[n]{a}=a^{\frac{1}{n}}$

b. The law of exponents for exponents applied to multiplication of terms in parentheses (in reverse direction):

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

Let's start by converting the square roots to exponents using the law of exponents mentioned in a:

$\sqrt{5}\cdot\sqrt{2}\cdot\sqrt{5}\cdot\sqrt{2}= \\ \downarrow\\ 5^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot5^{\frac{1}{2}}\cdot2^{\frac{1}{2}}=$

We'll continue, since there is multiplication between four terms with identical exponents, we can use the law of exponents mentioned in b (which also applies to multiplication of multiple terms in parentheses) and combine them together in multiplication under parentheses that are raised to the same exponent:

$5^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot5^{\frac{1}{2}}\cdot2^{\frac{1}{2}}= \\ (5\cdot2\cdot5\cdot2)^{\frac{1}{2}}=\\ 100^{\frac{1}{2}}=\\ \sqrt{100}=\\ \boxed{10}$

In the final steps, we first performed the multiplication within the parentheses, then we used again the root definition as an exponent mentioned in a (in reverse direction) to return to root notation, and in the final stage, we calculated the known square root of 100.

Therefore, we can identify that the correct answer is answer d.

To solve this problem, we'll simplify the given expression step by step:

First, let's simplify the numerator:

$\sqrt{10} \cdot \sqrt{5} \cdot \sqrt{2} = \sqrt{10 \cdot 5 \cdot 2} = \sqrt{100}$.

Simplifying further, $\sqrt{100} = 10$.

Next, simplify the denominator:

$\sqrt{5} \cdot \sqrt{5} \cdot \sqrt{4} = \sqrt{5 \cdot 5 \cdot 4} = \sqrt{100}$.

And $\sqrt{100} = 10$.

Now, divide the simplified numerator by the simplified denominator:

$\frac{10}{10} = 1$.

Therefore, the solution to the problem is $1$.

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

a. Root definition as an exponent:

$\sqrt[n]{a}=a^{\frac{1}{n}}$

b. The law of exponents for exponents applied to multiplication of terms in parentheses (in reverse order):

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

Let's start by converting the square roots to exponents using the law of exponents mentioned in a:

$\sqrt{5}\cdot\sqrt{10}\cdot\sqrt{2}\cdot\sqrt{4}= \\ \downarrow\\ 5^{\frac{1}{2}}\cdot10^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot4^{\frac{1}{2}}=$

We'll continue, since there is multiplication between four terms with identical exponents we can use the law of exponents mentioned in b (which also applies to multiplication of several terms in parentheses) and combine them together in multiplication under parentheses raised to the same exponent:

$5^{\frac{1}{2}}\cdot10^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot4^{\frac{1}{2}}=\\ (5\cdot10\cdot2\cdot4)^{\frac{1}{2}}=\\ 400^{\frac{1}{2}}=\\ \sqrt{400}=\\ \boxed{20}$

In the final stages, we first performed the multiplication within the parentheses, then we used again the root definition as an exponent mentioned earlier in a (in reverse order) to return to root notation, and in the final stage we calculated the known square root of 400.

Therefore, we can identify that the correct answer is answer c.

To solve the problem $\sqrt[3]{\sqrt{25}} \cdot \sqrt[3]{\sqrt{64}}$, we will work through it step by step:

Step 1: Simplify the inner square roots.

• $\sqrt{25}$ simplifies to $5$ because $5 \times 5 = 25$.
• $\sqrt{64}$ simplifies to $8$ because $8 \times 8 = 64$.

Step 2: Evaluate the cube roots.

• $\sqrt[3]{5}$ remains as $5^{1/3}$.
• $\sqrt[3]{8}$ evaluates to $2$ because $2 \times 2 \times 2 = 8$.

Step 3: Multiply the results of the cube roots.

• $5^{1/3} \cdot 2 = 2 \sqrt[3]{5}$.

Thus, the simplified expression is $2 \sqrt[3]{5}$.

Therefore, the solution to the problem is $2\sqrt[3]{5}$.

To solve the problem $\sqrt[5]{\sqrt{3}} \cdot \sqrt[5]{\sqrt{3}} =$, we follow these steps:

Step 1: Express each root using exponents.
$\sqrt[5]{\sqrt{3}}$ can be rewritten as $(3^{1/2})^{1/5}$, which simplifies to $3^{1/10}$ using the law $(a^m)^n = a^{m \cdot n}$.

Step 2: Multiply the expressions.
We have $(3^{1/10}) \cdot (3^{1/10})$. According to the laws of exponents, $a^m \cdot a^n = a^{m+n}$. Thus, the expression becomes $3^{1/10 + 1/10} = 3^{2/10} = 3^{1/5}$.

Step 3: Convert back to a root, if necessary.
The expression $3^{1/5}$ corresponds to $\sqrt[5]{3}$.

Therefore, the expression $\sqrt[5]{\sqrt{3}} \cdot \sqrt[5]{\sqrt{3}}$ simplifies to $3^{1/5}$, which is equivalent to $\sqrt[5]{9}$.

To match with the given choices, observe that $3^{1/5}$ can also be expressed as $\sqrt[10]{9}$ because $3^{1/5} = (3^2)^{1/10}$, which equals to $\sqrt[10]{9}$.

The correct answer is choice 4, $\sqrt[10]{9}$.