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2017 Paper 3 Q4
D: 1700.0 B: 1484.0
integration geometric mean logarithms exponential function change of base proof functional equation properties of integrals

For any function \(\f\) satisfying \(\f(x) > 0\), we define the geometric mean, F, by \[ F(y) = e^{\frac{1}{y} \int_{0}^{y} \ln f(x) \, dx} \quad (y > 0). \]

  1. The function f satisfies \(\f(x) > 0\) and \(a\) is a positive number with \(a\ne1\). Prove that \[ F(y) = a^{\frac{1}{y} \int_{0}^{y} \log_a f(x) \, dx}. \]
  2. The functions f and g satisfy \(\f(x) > 0\) and \(\g(x) > 0\), and the function \(\h\) is defined by \(\h(x) = \f(x)\g(x)\). Their geometric means are F, G and H, respectively. Show that \(H(y)= \F(y) \G(y)\,\).
  3. Prove that, for any positive number \(b\), the geometric mean of \(b^x\) is \(\sqrt{b^y}\,\).
  4. Prove that, if \(\f(x)>0\) and the geometric mean of \(\f(x)\) is \(\sqrt{\f(y)}\,\), then \(\f(x) = b^x\) for some positive number \(b\).

Solution:

  1. \begin{align*} && a^{\frac{1}{y} \int_{0}^{y} \log_a f(x) \, dx} &= e^{\ln a \cdot \frac{1}{y} \int_{0}^{y} \log_a f(x) \, dx} \\ &&&= e^{\ln a \cdot \frac{1}{y} \int_{0}^{y} \frac{\ln f(x)}{\ln a} \, dx} \\ &&&= e^{ \frac{1}{y} \int_{0}^{y} \ln f(x) \, dx} \\ &&&= F(y) \end{align*}
  2. \(\,\) \begin{align*} && H(y) &= e^{\frac1y \int_0^y \ln h(x) \d x} \\ &&&= e^{\frac1y \int_0^y \ln (f(x)g(x))\d x} \\ &&&= e^{\frac1y \int_0^y \left ( \ln f(x)+\ln g(x) \right)\d x} \\ &&&= e^{\frac1y \int_0^y \ln f(x) \d x +\frac1y \int_0^y \ln g(x) \d x} \\ &&&= e^{\frac1y \int_0^y \ln f(x) \d x }e^{\frac1y \int_0^y \ln g(x) \d x} \\ &&&= F(y)G(y) \end{align*}
  3. Suppose \(f(x) = b^x\), then \begin{align*} && F(y) &= b^{\frac1y \int_0^y \log_b f(x) \d x} \\ &&&= b^{\frac1y \int_0^y x \d x}\\ &&&= b^{\frac1y \frac{y^2}{2}} \\ &&&= b^{\frac{y}2} = \sqrt{b^y} \end{align*}
  4. Suppose the geometric mean of \(f(x)\) is \(\sqrt{f(y)}\) then the geometric mean of \(f(x)^2\) is \(f(y)\) by the the second part.

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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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1987 Paper 1 Q4
D: 1500.0 B: 1500.0
sequences infinite series geometric series logarithms change of base convergence summation

Show that the sum of the infinite series \[ \log_{2}\mathrm{e}-\log_{4}\mathrm{e}+\log_{16}\mathrm{e}-\ldots+(-1)^{n}\log_{2^{2^{n}}}\mathrm{e}+\ldots \] is \[ \frac{1}{\ln(2\sqrt{2})}. \] {[}\(\log_{a}b=c\) is equivalent to \(a^{c}=b\).{]}

Solution: Let \(S = \log_{2}\mathrm{e}-\log_{4}\mathrm{e}+\log_{16}\mathrm{e}-\ldots+(-1)^{n}\log_{2^{2^{n}}}\mathrm{e}+\ldots\) then \begin{align*} S &= \sum_{n=0}^{\infty} (-1)^n \log_{2^{2^n}} e \\ &= \sum_{n=0}^{\infty} (-1)^n \frac{\log e}{\log {2^{2^n}}} \\ &= \sum_{n=0}^{\infty} (-1)^n \frac{\log e}{2^n\log {2}} \\ &= \frac{\log e}{\log 2} \sum_{n=0}^{\infty} \frac{(-1)^n}{2^n} \\ &= \frac{1}{\log_e 2} \frac{1}{1+\frac12} \\ &= \frac{1}{\ln (2^{3/2})} \\ &= \frac{1}{\ln (2 \sqrt{2})} \end{align*}

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