Which is Greater? Exploring Exponential Expressions

When faced with the problem of determining which is greater between 98^{98} times 99^{99} and 100^{100}, it can be daunting. However, through a series of strategic mathematical transformations and simplifications, we can elucidate the answer with clarity.

Introduction

By breaking down each expression and applying logarithmic properties, we can simplify the problem and arrive at a definitive answer. In this article, we will explore the process in detail and discuss the underlying mathematical principles.

The Problem at Hand

We are tasked with comparing the values of:

98^{98} times 99^{99} 100^{100}

Strategic Simplification

To simplify the comparison, we will first express 100^{100} in a more manageable form:

100^{100} 100^{98} times 100^{2} 100^{98} times 100^{2}

Next, we will rewrite 98^{98} and 99^{99} in terms of 100:

98^{98} (100 - 2)^{98}

99^{99} (100 - 1)^{99}

By taking the natural logarithm of both expressions, we can further simplify our comparison:

(text{Let } A 98^{98} times 99^{99})

(text{Let } B 100^{100})

Now we take the natural logarithm of both A and B:

(ln A ln(98^{98}) ln(99^{99}) 98 ln 98 99 ln 99)

(ln B 100 ln 100)

Calculations and Comparisons

Let's calculate the exact values:

(ln 100 approx 4.60517) (ln 99 approx 4.59512) (ln 98 approx 4.58497)

(ln A approx 98 times 4.58497 99 times 4.59512)

(98 times 4.58497 approx 448.07406)

(99 times 4.59512 approx 454.76588)

(ln A approx 448.07406 454.76588 approx 902.83994)

(ln B approx 100 times 4.60517 approx 460.517)

By comparing the results, we see that:

(ln A approx 902.83994)

(ln B approx 460.517)

Since (ln A > ln B), we have:

(A > B)

Thus, we conclude that:

(98^{98} times 99^{99} > 100^{100})

Alternative Approach with Log Base 10

Another approach to solving this problem is by taking the logarithm base 10 of both expressions:

(log 98^{98} times 99^{99} log (98^{98}) log (99^{99}) 98 log 98 99 log 99 approx 392.7)

(log 100^{100} log 100^{100} 100 log 100 100 times 2 200)

When comparing these values, it becomes clear that:

(log 98^{98} times 99^{99} > log 100^{100})

Therefore, we confirm that:

(98^{98} times 99^{99} > 100^{100})

Conclusion

In conclusion, we have demonstrated multiple methods to show that 98^{98} times 99^{99} is indeed greater than 100^{100} by leveraging logarithmic properties and meticulous calculations. This problem not only improves our understanding of exponential expressions but also highlights the importance of logarithmic simplification in problem-solving.