Number Comparison: Identifying the Largest Value Among Given Numbers

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.