Problem 108 - Algebra and Trigonometry

Question

Distances Between Powers Which pair of numbers is closer together? $$ 10^{10} \text { and } 10^{50} \quad \text { or } \quad 10^{100} \text { and } 10^{101} $$

Step-by-Step Solution

Answer

10^{10} and 10^{50} are closer together.

Step 1: Understand the Problem

We need to compare the distance between the two pairs of numbers: (1) \(10^{10}\) and \(10^{50}\) and (2) \(10^{100}\) and \(10^{101}\). This involves calculating the difference between the numbers in each pair and seeing which difference is smaller.

Step 2: Calculate the Distance for the First Pair

The first pair is \(10^{10}\) and \(10^{50}\). Compute the distance between them as follows: The difference is \(10^{50} - 10^{10}\). Since \(10^{50}\) is much larger than \(10^{10}\), the difference is approximately equal to \(10^{50}\). Therefore, the distance is \(10^{50}\).

Step 3: Calculate the Distance for the Second Pair

The second pair is \(10^{100}\) and \(10^{101}\). Compute the distance between them as follows:The difference is \(10^{101} - 10^{100}\). We can factor out \(10^{100}\) from both terms:\[(10^{101} - 10^{100}) = 10^{100}(10 - 1) = 10^{100} \times 9 = 9 \times 10^{100}.\]Thus, the distance is \(9 \times 10^{100}\).

Step 4: Compare the Distances

Now, compare the distances computed in the previous steps. We have:- The distance for the first pair is \(10^{50}\).- The distance for the second pair is \(9 \times 10^{100}\).Clearly, \(10^{50} < 9 \times 10^{100}\) (because \(10^{100}\) is much larger than \(10^{50}\), and multiplying by 9 keeps it significantly larger).Thus, \(10^{10}\) and \(10^{50}\) are closer together than \(10^{100}\) and \(10^{101}\).

Key Concepts

Distance between Powers

The concept of the distance between two powers involves understanding how far apart two exponential numbers are on a number line. When we talk about distance in this context, we mean the difference between the values of these numbers. Consider the two pairs given:

• The first pair: \(10^{10}\) and \(10^{50}\)
• The second pair: \(10^{100}\) and \(10^{101}\)

To find out which pair of numbers is closer, we calculate the distance by subtracting the smaller power from the larger power in each pair.
- For the first pair, the difference is \(10^{50} - 10^{10}\), which simplifies to approximately \(10^{50}\) since \(10^{50}\) is much greater than \(10^{10}\).
- For the second pair, the subtraction \(10^{101} - 10^{100}\) simplifies to \(9 \times 10^{100}\), because factoring out \(10^{100}\) gives \(10^{101} - 10^{100} = 10^{100}(10 - 1)\).
Comparing these distances helps us understand which pair of numbers is closer together.

Comparison of Large Numbers

When comparing large numbers, especially those expressed as powers of ten, it is crucial to understand the magnitude depicted by each number. Large powers of ten are significantly greater than smaller ones, thus impacting the calculation of their distances.
In scientific notation, the powers represent the number of zeros following the 1 in the number. For instance:

• \(10^{50}\) is a 1 followed by 50 zeros.
• \(10^{100}\) is a 1 followed by 100 zeros.

In the context of this exercise:
- The first pair \(10^{50}\) and \(10^{10}\) results in a significant distance because \(10^{50}\) dwarfs \(10^{10}\).
- The second pair \(10^{100}\) and \(10^{101}\) results in a lesser disparity relative to each other in the context of their size, despite \(10^{101}\) being just ten times more than \(10^{100}\).
Recognizing which numbers are exponentially larger aids in accurate calculation and understanding of relationships between vast magnitudes.

Mathematical Subtraction

Subtraction, when applied to large numbers and exponents, involves considering the not just the numerical values but the exponents that denote their magnitude.
The operation \(a^n - a^m\) can often be simplified if the exponents are closely related. Here is how it works in the context of our exercise:

• For the first pair, \(10^{50} - 10^{10}\) results in a difference close to \(10^{50}\), emphasizing how much larger \(10^{50}\) is compared to \(10^{10}\).
• For the second pair, \(10^{101} - 10^{100}\), we recognize this as \((10^{100}) \times (10 - 1) = 9 \times 10^{100}\).

Understanding subtraction here involves factoring and recognizing how powers of ten multiply.
This knowledge helps to tackle challenges related to distances and differences between powers, especially when those powers are vast and not immediately comparable through regular calculation methods.