Solution: \(x = \log_bc\) means that \(b^x = c\)
Therefore, we can write \(\frac{\log_ac}{\log_ab} = \frac{\log_ab^{x}}{\log_ab} = \frac{x \log_ab}{\log_ab} = x = \log_bc\), giving us the change of base rule.
Rearranging the chance of base rule, we get \(\frac{1}{\log_bc} = \frac{\log_ab}{\log_ac}\)
- Since \(\pi^2 < 10\), we have
\begin{align*}
\Leftrightarrow && \pi^2 &< 10 \\
\Leftrightarrow && 2\log_{10} \pi &< 1 \\
\Leftrightarrow && 2 &< \frac{1}{\log_{10} \pi} \\
\Leftrightarrow && 2 &< \frac{\log_{10} 2 + \log_{10} 5}{\log_{10} \pi} \\
\Leftrightarrow && 2 &< \frac{\log_{10} 2}{\log_{10} \pi} + \frac{\log_{10} 5}{\log_{10} \pi} \\
\Leftrightarrow && 2 &< \frac1{\log_2\pi}+ \frac1{\log_5\pi} \\
\end{align*}
- Since \(e^2 < 8 \Rightarrow 2 < 3 \ln 2 \Rightarrow \ln 2 > \frac23\)
\begin{align*}
&& \log_2 \frac{\pi}{e} &= \frac{\ln \frac{\pi}{e}}{\ln 2} \\
&&&= \frac{\ln \frac{\pi}{e}}{\ln 2} \\
&&&< \frac{3(\ln \pi - 1)}{2} \\
\Rightarrow && \frac{1}{5} &< \frac{3 \ln \pi - 1}{2} \\
\Rightarrow && \frac{2}{15} + 1 = \frac{17}{15}&< \ln \pi
\end{align*}
- From the inequalities given we can set up two linear inequalities in \(\ln 2\) and \(\ln 5\)
\begin{align*}
&& 20 &< e^3 \\
\Rightarrow && 2\ln 2 + \ln 5 &< 3 \tag{*}\\
\\
&& \frac{3}{10} &< \log_{10} 2 \\
\Rightarrow && \frac{3}{10} &< \frac{\ln 2}{\ln 10} \\
\Rightarrow && 3 \ln 2 + 3 \ln 5 &< 10 \ln 2 \\
\Rightarrow && -7 \ln 2 + 3 \ln 5 &< 0 \tag{**}\\
\\
\\
&& \pi^2 & < 10 \\
\Rightarrow && 2 \ln \pi &< \ln 2 + \ln 5 \tag{***}\\
\end{align*}
It would be nice to combine \((*)\) and \((**)\) to bound \(\ln 2+ \ln 5\), so we want to use a linear combination such that
\begin{align*}
&& \begin{cases} 2x -7y &= 1 \\
x + 3y &= 1\end{cases} \\
\Rightarrow && \begin{cases} y &= \frac1{13} \\
x &= \frac{10}{13}\end{cases} \\
\\
\Rightarrow && \ln 2 + \ln 5 < \frac{30}{13} + 0 \\
\Rightarrow && \ln \pi < \frac{15}{13}
\end{align*}