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2000 Paper 1 Q1
D: 1500.0 B: 1599.6
logarithms change of base powers first digit inequality log10 estimation

To nine decimal places, \(\log_{10}2=0.301029996\) and \(\log_{10}3=0.477121255\).

  1. Calculate \(\log_{10}5\) and \(\log_{10}6\) to three decimal places. By taking logs, or otherwise, show that \[ 5\times 10^{47} < 3^{100} < 6\times 10^{47}. \] Hence write down the first digit of \(3^{100}\).
  2. Find the first digit of each of the following numbers: \(2^{1000}\); \ \(2^{10\,000}\); \ and \(2^{100\, 000}\).

Solution:

  1. \begin{align*} \log_{10}5 &= \log_{10} 10 - \log_{10}2 \\ &= 1- \log_{10} 2 \\ &= 0.699\\ \\ \log_{10} 6 &= \log_{10} 2 + \log_{10} 3 \\ &= 0.301029996+0.477121255 \\ &= 0.778 \end{align*} \begin{align*} && 5 \times 10^{47} < 3^{100} < 6 \times 10^{47} \\ \Leftrightarrow && 47 + \log_{10} 5 < 100 \log_{10} 3 < \log_{10} 6 + 47 \\ \Leftrightarrow &&47.699< 47.71 < 47.778 \\ \end{align*} Which is true. Therefore the first digit of \(3^{100}\) is 5.
  2. \(\log_{10} 2^{1000} = 1000 \log_{10} 2 = 301.02\cdots\). Therefore it starts with a \(1\). \(\log_{10}2^{10\, 000} = 10\,000 \log_{10} 2 = 3010.2\) therefore this also starts with a \(1\). \(\log_{10} 2^{100\, 000} = 100\,000 \log_{10} 2 = 30102.9996\) therefore it starts with a \(9\)

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