Number Comparison: Identifying the Largest Value Among Given Numbers

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.