Question

If \(a\) is a 100 -digit number, is the number of digits of \(a^{10^{120}}\) closer to \(10^{120}\) or \(10^{240}\) ? Is it a lot closer? Does the answer depend on what \(a\) actually is rather than the number of digits it has?

Short answer

The number of digits is much closer to \(10^{120}\) regardless of the specific 100-digit value of \(a\).

Step-by-step solution

Understand the Problem

We need to determine whether the number of digits in \(a^{10^{120}}\) is closer to \(10^{120}\) or \(10^{240}\). The number \(a\) is a 100-digit number, meaning it ranges from \(10^{99}\) to \(10^{100}-1\). We must explore how the number of digits of \(a^{10^{120}}\) compares to both given powers of 10.

Determine the Number of Digits in the Expression

To find the number of digits of \(a^{10^{120}}\), we use the formula for the number of digits of a number \(n\), which is \(\lfloor \log_{10} n \rfloor + 1\). Thus, the number of digits in \(a^{10^{120}}\) is \(\lfloor \log_{10} (a^{10^{120}}) \rfloor + 1\).

Apply Logarithm Properties

Using the logarithm property \(\log (x^y) = y \log x\), we find that \(\log_{10} (a^{10^{120}}) = 10^{120} \cdot \log_{10} a\). Therefore, the number of digits is \(\lfloor 10^{120} \cdot \log_{10} a \rfloor + 1\).

Estimate \(\log_{10} a\) Range

Since \(a\) is a 100-digit number, \(a\) is approximately \(10^{99} \leq a < 10^{100}\). Thus, \(99 \leq \log_{10} a < 100\). This means \(10^{120} \cdot \log_{10} a\) ranges from \(10^{120} \times 99\) to \(10^{120} \times 100\), i.e., \(99 \times 10^{120}\) to \(100 \times 10^{120}\).

Compare the Result to Given Powers

We now compare this range to \(10^{120}\) and \(10^{240}\). Since both bounds \(99 \times 10^{120}\) and \(100 \times 10^{120}\) fall within the magnitudes of \(10^{121}\) to \(10^{122}\), which are very close to \(10^{120}\), it shows the result is much closer to \(10^{120}\) than to \(10^{240}\).

Conclusion

Since the power of 10 to which the number of digits matches is \(10^{120} \), and given the ranges, the proximity is substantially closer to \(10^{120}\). Additionally, the slight variation in \(a\) being within 100-digit numbers does not significantly affect this proximity.

Key concepts

Logarithms

Logarithms are a powerful mathematical tool that help us work with large numbers. Essentially, a logarithm tells us how many times a particular number, called the base, must be multiplied by itself to equal a given number. For example, the logarithm base 10 of 100 is 2 because 10 multiplied by itself twice equals 100.

Logarithms are closely tied to exponentiation, which is raising a number to a power. One of the main properties we use is that the logarithm of a power is the exponent times the logarithm of the base: \( \log_b(x^y) = y \cdot \log_b(x) \). This property is particularly useful when we need to calculate the number of digits in a large number.

• Logarithms simplify the process of multiplication into addition.
• They help compare very large numbers by putting them on a "logarithmic scale", which is much easier to handle.
• Using logarithms, we can transform exponential comparisons into simple arithmetic.

In problems involving large exponents and comparisons as in the exercise, logarithms are essential to determine which large number is "closer" in terms of its digit count.

Number of digits

The number of digits in a number gives us a sense of its size. If you think about digit count, it reflects how many numerical places a number takes up. To find the number of digits in any number \( n \), you can use the formula: \( \lfloor \log_{10} n \rfloor + 1 \). This is because the base-10 logarithm of a number indicates its magnitude in powers of 10.

Consider the relation for our problem: the number of digits in \( a^{10^{120}} \). By applying the properties of logarithms, we can express this as \( \lfloor 10^{120} \cdot \log_{10} a \rfloor + 1 \). Here’s why:

• A logarithm helps convert the exponential growth into something tangible—like digit count.
• When \( a \) is a 100-digit number (meaning \( 10^{99} \leq a < 10^{100} \)), its logarithm gives us a linear range between 99 and 100.
• As the base (\( a \)) grows with each digit added, the result of exponentiation will produce significantly more digits very quickly.

In practical terms, knowing how to calculate the number of digits helps us understand just how large an exponential number like \( a^{10^{120}} \) truly is.

Mathematical comparison

Comparing large numbers, especially those involving exponentiation, can be tricky. It requires a clear understanding of magnitudes and scales. When our exercise asks if the number of digits in \( a^{10^{120}} \) is closer to \( 10^{120} \) or \( 10^{240} \), it's all about determining which quantity is larger when judged on a relative scale.

The comparison process involves:

• First, estimating the number of digits of \( a^{10^{120}} \) using logarithms.
• Then, analyzing how these digits align with \( 10^{120} \) and \( 10^{240} \).
• Finally, assessing the closeness of those digits to each benchmark.

For instance, in our problem, we find the digit range by using \( 10^{120} \times 99 \) to \( 10^{120} \times 100 \)—a range from \( 99 \times 10^{120} \) to \( 100 \times 10^{120} \). This range is much closer to \( 10^{120} \) than \( 10^{240} \).

Mathematical comparison in this context focuses on magnitude over raw calculations, helping us discern relative size and "closeness" more effectively than mere numerical difference could.