Product Property of Square Roots: Identify the greater value

To determine which of the suggested options has the largest numerical value, we will use two laws of exponents:

a. Definition of root as an exponent:

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

b. Law of exponents for exponents in parentheses (in reverse direction):

$a^n\cdot b^n=(a\cdot b)^n$

We will therefore deal with options a and c first (in the answers), starting by converting the square root to exponent notation, using the law of exponents mentioned in a earlier:

$$$\sqrt{2}\cdot\sqrt{3} \rightarrow 2^{\frac{1}{2}}\cdot3^{\frac{1}{2}}\\ \sqrt{1}\cdot\sqrt{6} \rightarrow 1^{\frac{1}{2}}\cdot6^{\frac{1}{2}}\\$We will continue, since both terms in the multiplication (in both options we are currently dealing with) have the same exponent, we can use the law of exponents mentioned in b earlier to combine them in parentheses which are raised to the same exponent and then calculate the result of the multiplication in parentheses:

$2^{\frac{1}{2}}\cdot3^{\frac{1}{2}} \rightarrow (2\cdot3)^{\frac{1}{2}}=6^{\frac{1}{2}} \\ 1^{\frac{1}{2}}\cdot6^{\frac{1}{2}}\rightarrow(1\cdot6)^{\frac{1}{2}}=6^{\frac{1}{2}} \\$In the next step, we will return to root notation, again, using the law of exponents mentioned in a (in reverse direction):

$6^{\frac{1}{2}}\rightarrow\sqrt{6} \\$We can identify now that the numerical values of options a, b, and c are equal, since we got that:

$\sqrt{2}\cdot\sqrt{3} \rightarrow 2^{\frac{1}{2}}\cdot3^{\frac{1}{2}}=6^{\frac{1}{2}}=\sqrt{6}\\ \sqrt{1}\cdot\sqrt{6} \rightarrow 1^{\frac{1}{2}}\cdot6^{\frac{1}{2}}=6^{\frac{1}{2}}=\sqrt{6}\\$Therefore, we need to determine which of these expressions:

$\sqrt{6}, \hspace{6pt}\sqrt{9}$has a higher numerical value,

We can determine this by converting these two values to exponent notation, again, using the law of exponents mentioned in a:

$\sqrt{6}\rightarrow6^{\frac{1}{2}}\\ \sqrt{9}\rightarrow9^{\frac{1}{2}}\\$Note that these two expressions have the same exponent (and their bases are positive, we'll mention, although obvious), therefore we can determine their relationship by comparing only their bases, since it will be identical:

9>6\hspace{4pt} (>0)\\ \downarrow\\ 9^{\frac{1}{2}}>6^{\frac{1}{2}} In other words, we got that:

\sqrt{9}>\sqrt{6}

Therefore, the correct answer is answer d.

To determine which of the suggested options has the largest numerical value, we will use the definition of root as a power:

$\sqrt[n]{a}=a^{\frac{1}{n}}$Let's substitute each one of the square roots in the suggested options with powers:

$\sqrt{2}\rightarrow2^{\frac{1}{2}}\\ \sqrt{3}\rightarrow3^{\frac{1}{2}}\\ \sqrt{4}\rightarrow4^{\frac{1}{2}}\\ \sqrt{5}\rightarrow5^{\frac{1}{2}}\\$Now let's note that all the expressions we got have the same exponent (and their bases are positive, we'll also mention, although it's obvious), therefore we can determine the trend between them using only the trend between their bases, since it's identical to it:

5>4>3>2\hspace{4pt} (>0)\\ \downarrow\\ 5^{\frac{1}{2}}>4^{\frac{1}{2}} >3^{\frac{1}{2}}>2^{\frac{1}{2}} In other words, we got that:

\sqrt{5}>\sqrt{4}>\sqrt{3}>\sqrt{2} Therefore the correct answer is answer D.

To determine which of the suggested options has the largest numerical value, we will use two laws of exponents:

a. Definition of root as an exponent:

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

b. Law of exponents for an exponent applied to terms in parentheses (in reverse order):

$a^n\cdot b^n=(a\cdot b)^n$

Let's start by converting the square root in each of the suggested options (except D) to exponential notation, using the law of exponents mentioned in a above:

$$$\sqrt{36}\cdot\sqrt{1} \rightarrow 36^{\frac{1}{2}}\cdot1^{\frac{1}{2}}\\ \sqrt{6}\cdot\sqrt{6} \rightarrow 6^{\frac{1}{2}}\cdot6^{\frac{1}{2}}\\ \sqrt{9}\cdot\sqrt{4} \rightarrow 9^{\frac{1}{2}}\cdot4^{\frac{1}{2}}\\$We'll continue, since both terms in the multiplication (in all the options we're dealing with now) have the same exponent, we can use the law of exponents mentioned in b above and combine them together in the multiplication within parentheses which are raised to the same exponent:

$36^{\frac{1}{2}}\cdot1^{\frac{1}{2}} \rightarrow (36\cdot1)^{\frac{1}{2}}=36^{\frac{1}{2}} \\ 6^{\frac{1}{2}}\cdot6^{\frac{1}{2}}\rightarrow(6\cdot6)^{\frac{1}{2}}=36^{\frac{1}{2}} \\ 9^{\frac{1}{2}}\cdot4^{\frac{1}{2}} \rightarrow (9\cdot4)^{\frac{1}{2}}=36^{\frac{1}{2}} \\$Let's summarize what we've done so far, we got that:

$\sqrt{36}\cdot\sqrt{1}=36^{\frac{1}{2}}\\ \sqrt{6}\cdot\sqrt{6}= 36^{\frac{1}{2}}\\ \sqrt{9}\cdot\sqrt{4}= 36^{\frac{1}{2}}\\$Note now that the values of all expressions suggested in options A-C are equal to each other.

Therefore, the correct answer is D.

To determine which of the suggested options has the largest numerical value, we will use two laws of exponents:

a. Definition of root as an exponent:

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

b. Law of exponents for exponents applied to expressions in parentheses (in reverse direction):

$a^n\cdot b^n=(a\cdot b)^n$

Let's deal with options a and b (in the answers) first, we'll start by converting the square root to exponent notation, using the law of exponents mentioned in a earlier:

$$$\sqrt{4}\cdot\sqrt{9} \rightarrow 4^{\frac{1}{2}}\cdot9^{\frac{1}{2}}\\ \sqrt{4}\cdot\sqrt{1} \rightarrow 4^{\frac{1}{2}}\cdot1^{\frac{1}{2}}\\$We'll continue, since both terms in the multiplication (in both options we're currently dealing with) have the same exponent, we can use the law of exponents mentioned in b earlier and combine them together in a multiplication within parentheses raised to the same exponent and then calculate the result of the multiplication in parentheses:

$4^{\frac{1}{2}}\cdot9^{\frac{1}{2}} \rightarrow (4\cdot9)^{\frac{1}{2}}=36^{\frac{1}{2}} \\ 4^{\frac{1}{2}}\cdot1^{\frac{1}{2}}\rightarrow(4\cdot1)^{\frac{1}{2}}=4^{\frac{1}{2}} \\$In the next step, we'll return to root notation, again, using the law of exponents mentioned in a (in reverse direction) and then use the (known) roots of the numbers that will be obtained in the root:

$36^{\frac{1}{2}}\rightarrow\sqrt{36}=6 \\ 4^{\frac{1}{2}}\rightarrow\sqrt{4}=2 \\$We have thus found that the number in option a is larger since:

6>2

Additionally, we identify that the numerical values of options a and c are equal.

Therefore, the correct answer is answer d.

To determine which of the suggested options has the largest numerical value, we will use two laws of exponents:

a. Definition of root as an exponent:

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

b. Law of exponents for exponents in parentheses (in reverse order):

$a^n\cdot b^n=(a\cdot b)^n$

Let's start by converting the fourth root in each of the suggested options to exponent notation, using the law of exponents mentioned in a above:

$$$\sqrt{2}\cdot\sqrt{1} \rightarrow 2^{\frac{1}{2}}\cdot1^{\frac{1}{2}}\\ \sqrt{2}\cdot\sqrt{2} \rightarrow 2^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\\ \sqrt{2}\cdot\sqrt{3} \rightarrow 2^{\frac{1}{2}}\cdot3^{\frac{1}{2}}\\ \sqrt{2}\cdot\sqrt{4} \rightarrow 2^{\frac{1}{2}}\cdot4^{\frac{1}{2}}\\$We'll continue, since both terms in the multiplication (in all the options we're dealing with now) have the same exponent, we can use the law of exponents mentioned in b above and combine them together in the multiplication within parentheses raised to the same exponent, and then calculate the result of the multiplication in parentheses:

$2^{\frac{1}{2}}\cdot2^{\frac{1}{2}} \rightarrow (2\cdot1)^{\frac{1}{2}}=2^{\frac{1}{2}} \\ 2^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\rightarrow(2\cdot2)^{\frac{1}{2}}=4^{\frac{1}{2}} \\ 2^{\frac{1}{2}}\cdot3^{\frac{1}{2}} \rightarrow (2\cdot3)^{\frac{1}{2}}=6^{\frac{1}{2}} \\ 2^{\frac{1}{2}}\cdot4^{\frac{1}{2}}\rightarrow(2\cdot4)^{\frac{1}{2}}=8^{\frac{1}{2}} \\$Let's summarize what we've done so far, we got that:

$\sqrt{2}\cdot\sqrt{1}=2^{\frac{1}{2}}\\ \sqrt{2}\cdot\sqrt{2}= 4^{\frac{1}{2}}\\ \sqrt{2}\cdot\sqrt{3}= 6^{\frac{1}{2}}\\ \sqrt{2}\cdot\sqrt{4}= 8^{\frac{1}{2}}\\$Now let's note that all the expressions we got have the same exponent (and their bases are positive, we'll mention, although it's obvious), therefore we can determine the trend between them using only the trend between their bases, since it's identical to it:

8>6>4>2\hspace{4pt} (>0)\\ \downarrow\\ 8^{\frac{1}{2}}>6^{\frac{1}{2}} >4^{\frac{1}{2}}>2^{\frac{1}{2}}

Therefore the correct answer is answer d.

To determine which of the suggested options has the largest numerical value, we will use the following root law:

$\sqrt[\textcolor{blue}{n}]{a^{\textcolor{red}{m}}}=a^{\frac{\textcolor{red}{m}}{\textcolor{blue}{n}}} =(\sqrt[\textcolor{blue}{n}]{a})^{\textcolor{red}{m}}$

Let's start by applying this law to each of the suggested options (and remember that a square root is a second-order root - which we don't explicitly write next to the root), meaning - we will convert the roots to exponential notation, then we'll use the (known) root of the number 25:

$\sqrt{25^{\textcolor{red}{2}}}=25^{\frac{\textcolor{red}{2}}{\textcolor{blue}{2}}}=(\sqrt{25})^{\textcolor{red}{2}}=5^2 \\ \sqrt{25}=\sqrt{25^{\textcolor{red}{1}}}=25^{\frac{\textcolor{red}{1}}{\textcolor{blue}{2}}}=(\sqrt{25})^{\textcolor{red}{1}}=5^1 \\ \sqrt{25^{\textcolor{red}{3}}}=25^{\frac{\textcolor{red}{3}}{\textcolor{blue}{2}}}=(\sqrt{25})^{\textcolor{red}{3}}=5^3 \\ \sqrt{25^{\textcolor{red}{4}}}=25^{\frac{\textcolor{red}{4}}{\textcolor{blue}{2}}}=(\sqrt{25})^{\textcolor{red}{4}}=5^4 \\$We got four options which are all powers of the same base (5), since this base is greater than 1, the largest option will be the one where the base (5) is raised to the highest power (and the opposite if the base is between 0 and 1, then as the power increases the value of the number - meaning the base raised to that power - decreases),

Therefore:

5^4>5^3>5^2>5^1

Thus answer D is the correct answer.

To determine which of the suggested options has the largest numerical value, we will use three laws of exponents:

a. Definition of root as an exponent:

$\sqrt[n]{a}=a^{\frac{1}{n}}$b. Law of exponents for an exponent applied to a product in parentheses (in reverse order):

$a^n\cdot b^n=(a\cdot b)^n$c. Law of exponents for an exponent raised to an exponent:

$(a^m)^n=a^{m\cdot n}$Let's deal with each of the suggested options (in the answers), starting by converting the square root to exponent notation, using the law of exponents mentioned in a' earlier:

$$$\sqrt{5}\cdot\sqrt{5} \rightarrow 5^{\frac{1}{2}}\cdot5^{\frac{1}{2}}\\ \sqrt{2}\cdot\sqrt{2} \rightarrow 2^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\\ \sqrt{3}\cdot\sqrt{3} \rightarrow 3^{\frac{1}{2}}\cdot3^{\frac{1}{2}}\\ \sqrt{4}\cdot\sqrt{4} \rightarrow 4^{\frac{1}{2}}\cdot4^{\frac{1}{2}}\\$Let's continue, since both terms in the product (in both options we are currently dealing with) have the same exponent, we can use the law of exponents mentioned in b' earlier and combine them in the parentheses product raised to the same exponent and then calculate the result of the product in parentheses:

$5^{\frac{1}{2}}\cdot5^{\frac{1}{2}} \rightarrow (5\cdot5)^{\frac{1}{2}}=(5^2)^{\frac{1}{2}} \\ 2^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\rightarrow(2\cdot2)^{\frac{1}{2}}=(2^2)^{\frac{1}{2}} \\ 3^{\frac{1}{2}}\cdot3^{\frac{1}{2}} \rightarrow (3\cdot3)^{\frac{1}{2}}=(3^2)^{\frac{1}{2}} \\ 4^{\frac{1}{2}}\cdot4^{\frac{1}{2}}\rightarrow(4\cdot4)^{\frac{1}{2}}=(4^2)^{\frac{1}{2}} \\$Let's continue, we'll apply the law of exponents mentioned in c' and calculate the exponent on the term (with exponent) in parentheses:

$(5^2)^{\frac{1}{2}}\rightarrow 5^{2\cdot \frac{1}{2}}=5^1=5 \\ (2^2)^{\frac{1}{2}}\rightarrow 2^{2\cdot \frac{1}{2}}=2^1=2 \\ (4^2)^{\frac{1}{2}}\rightarrow 4^{2\cdot \frac{1}{2}}=4^1=4 \\ (3^2)^{\frac{1}{2}}\rightarrow 3^{2\cdot \frac{1}{2}}=3^1=3 \\$

We have therefore found that the number in option a' is the largest because:

5>4>3>2 Therefore, the correct answer is answer a'.