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Chapter 7: Problem 14

In the decimal expansion of \(0.9^{9999}\), how many zeros follow the decimal point before the first nonzero digit?

Short Answer

Expert verified
In the decimal expansion of \(0.9^{9999}\), there are 461 zeros following the decimal point before the first nonzero digit.

Step by step solution

01

Understand Decimal Expansion in Powers

When we raise a number between 0 and 1 to a power, its decimal expansion shrinks. In this case, as 0.9 is close to 1, the shrinking will be minimal. Nonetheless, the exponent is large (9999), which means that even minimal shrinking can produce a significant number of zeros.
02

Write the Decimal Expansion of \(0.9^{9999}\) as an Inequality

To find the position of the first nonzero digit, we will consider which powers of 10 can bound \(0.9^{9999}\) as an inequality. If \(10^{-n} < 0.9^{9999} < 10^{-n+1}\), then \(0.9^{9999}\) can be expressed as \(0.\underbrace{00\cdots 0}_{n \text{ zeros}}x\), where \(x\) is nonzero.
03

Solve the Inequality Using Logarithms

Taking the logarithm base 10 of the inequality, we get \(-n < \log_{10}(0.9^{9999}) < -n + 1\). As \(0.9 = \frac{9}{10}\), we can rewrite, \(\log_{10}(0.9) = \log_{10}(9) - \log_{10}(10)\) or, \(\log_{10}(9) = 1 - \log_{10}(10)\). Now, the inequality can be expressed as \(-n < 9999 (\log_{10}(9) - 1) < -n + 1\). Let's find the integer value for n that satisfies this inequality.
04

Determine the Integer Value of n

-n < 9999 (\(\log_{10}(9)\) - 1) < -n + 1 \(\implies\) -n < 9999 \(\log_{10}(9)\) - 9999 < -n + 1 \(\implies\) n > 9999 - 9999 \(\log_{10}(9)\) > n - 1 Using a calculator, we can find that \(\log_{10}(9) \approx 0.954\). Thus, n > 9999 - 9999(0.954) > n - 1 \(\implies\) n > 460 > n - 1 This inequality is satisfied by n = 461.
05

Summarize the Results

The integer value of n that satisfies the inequality is 461. In the decimal expansion of \(0.9^{9999}\), there are 461 zeros following the decimal point before the first nonzero digit.

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