Comparing Values with Condition a > 1: Finding the Maximum

Note that in all options there are fractions where both numerator and denominator have terms with identical bases, therefore we will use the division law between terms with identical bases to solve the exercise:

$\frac{c^m}{c^n}=c^{m-n}$Let's apply it to the problem, first we'll simplify each of the suggested options using the above law (options in order):

$\frac{a^{17}}{a^{20}}=a^{17-20}=a^{-3}$$\frac{a^{10}}{a^1}=a^{10-1}=a^9$$\frac{a^3}{a^{-2}}=a^{3-(-2)}=a^{3+2}=a^5$$\frac{a^2}{a^4}=a^{2-4}=a^{-2}$Let's return to the problem, given that:

$a>1$therefore the option with the largest value will be the one where $a$has the largest exponent (for emphasis - a positive exponent is greater than a negative exponent),

meaning the option:

$a^9$above is correct, it came from option B in the answers,

therefore answer B is correct.