Rules of Roots Combined: Identify the greater value

Let's begin by calculating the numerical value of each of the roots in the given options:

$\sqrt{25}=5\\ \sqrt{16}=4\\ \sqrt{9}=3\\$We can determine that:

5>4>3>1 Therefore, the correct answer is option A

In order to determine which of the suggested options has the largest numerical value, we will apply 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):

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

Let's proceed to examine the options a and c (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}}\\$Due to the fact that both terms in the multiplication have the same exponent, we are able to apply the law of exponents mentioned in b to combine them inside of parentheses, which are subsequently raised to the same exponent. Once completed proceed to calculate the result of the multiplication inside of the 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 deduce that the numerical values of options a, b, and c are equal, as seen below:

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

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

This can be achieved 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), Therefore we can determine their relationship by simply comparing 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.

In order to determine which of the given options has the largest numerical value, apply 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 begin by examining options a and b (in the answers)

Convert 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}}\\$Due to the fact that both terms in the multiplication have the same exponent, we can use the law of exponents mentioned in b earlier proceed to combine them together in a multiplication operation within parentheses, whilst raised to the same exponent . Once completed calculate the result of the multiplication inside of the 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, return to root notation, again, using the law of exponents mentioned in a (in reverse direction) and then apply 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 given that:

6>2

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

Therefore, the correct answer is answer d.

In order to determine which of the suggested options has the largest numerical value, we will apply 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 obtained 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.

In order to determine which of the following options has the largest numerical value, we will apply 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}}\\$Due to the fact that both terms in the multiplication have the same exponent, we are able to apply the law of exponents mentioned in b above and combine them together in the multiplication within parentheses, whilst raised to the same exponent. Once completed we can then calculate the result of the multiplication inside of the 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, as shown below:

$\sqrt{2}\cdot\sqrt{1}=2^{\frac{1}{2}}\\ \sqrt{2}\cdot\sqrt{2}= 4^{\frac{1}{2}}\\ \sqrt{2}\cdot\sqrt{3}= 6^{\frac{1}{2}}\\ \sqrt{2}\cdot\sqrt{4}= 8^{\frac{1}{2}}\\$Now let's note that all the expressions we obtained have the same exponent (they're bases are also positive), 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.

In order to determine which of the following options has the largest numerical value, we will apply 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}}\\$Given that both terms in the multiplication 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 subsequently 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, as shown below:

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

Therefore, the correct answer is D.

In order to determine which of the suggested options has the largest numerical value, apply the 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}}\\$Due to the fact that both terms in the product have the same exponent, we are able to apply the law of exponents mentioned in b' earlier and then proceed to combine them together inside of the parentheses product, raised to the same exponent . Once completed we can then calculate the result of the product in the parentheses:

$5^{\frac{1}{2}}\cdot5^{\frac{1}{2}} \rightarrow (5\cdot5)^{\frac{1}{2}}=(5^2)^{\frac{1}{2}} \\ 2^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\rightarrow(2\cdot2)^{\frac{1}{2}}=(2^2)^{\frac{1}{2}} \\ 3^{\frac{1}{2}}\cdot3^{\frac{1}{2}} \rightarrow (3\cdot3)^{\frac{1}{2}}=(3^2)^{\frac{1}{2}} \\ 4^{\frac{1}{2}}\cdot4^{\frac{1}{2}}\rightarrow(4\cdot4)^{\frac{1}{2}}=(4^2)^{\frac{1}{2}} \\$Proceed to apply the law of exponents mentioned in c' and then calculate the exponent inside of the 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 determined that the number in option a' is representative of the largest value:

5>4>3>2 The correct answer is a'.