Numerical Comparison: Identify the Maximum 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 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.