Value Comparison: Identifying the Maximum in a Number Set

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.