Question

Arrange the following in descending order
(A) \(3^{3^{33}}\)
(B) \(3^{(33)^3}\)
(C) \((3^3)^{33}\)
(D) \(3^{333}\)
Choose the correct answer from the options given below :

• A. \(A>B>D>C\)
• B. \(B = C>A>D\)
• C. \(B>A>D>C\)
• D. \(A>C>D>B\)

Hint:

Power towers grow much faster than any other exponential form. In $a^{b^c}$, $c$ is at the highest level and has the most impact. Always evaluate exponents in tower form from top to bottom.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For comparison of large numbers with the same base, we must simplify and compare the exponents. Note the difference in notation: \(a^{b^c} = a^{(b^c)}\) (evaluated top-down), whereas \((a^b)^c = a^{b \times c}\).
Step 2: Key Formula or Approach: Compare the magnitude of exponents for base 3.
Step 3: Detailed Explanation:
Let's analyze the exponent for each term:
(A) \(3^{3^{33}}\): The exponent is \(3^{33}\). This is a massive number, approximately \(5.5 \times 10^{15}\).
(B) \(3^{(33)^3}\): The exponent is \(33^3 = 35,937\).
(C) \((3^3)^{33} = 3^{3 \times 33} = 3^{99}\). The exponent is \(99\).
(D) \(3^{333}\): The exponent is \(333\).
Now, we compare the exponents:
\(3^{33}\) is significantly larger than \(35,937\).
\(35,937\) is significantly larger than \(333\).
\(333\) is larger than \(99\).
Therefore, the order of exponents is: \(3^{33}>(33)^3>333>99\).
Correspondingly, the order of the numbers is \(A>B>D>C\).
Step 4: Final Answer:
The descending order is \(A>B>D>C\).