log question (1 Viewer)
- Thread starter Mahan1
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If
then find a.
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well, I used your method, except I chose to work with base a. But I like challenge and as you suggested, there might be a faster method. I tried to find a different or possibly faster method, and this is what I got: \\ \begin{align*} \log_a{16} + \log_{\sqrt{2}}a &= 9 \\ \frac{\log_{16}{16}}{\log_{16}{a}} + \frac{\log_{16}{a}}{\log_{16}{\sqrt{2}}} &= 9 \\ \frac{1}{\log_{16}{a}} + \frac{\log_{16}{a}}{\frac{1}{8}} &= 9 \\ \frac{1}{\log_{16}{a}} + 8\log_{16}{a} &=9 \\ 1+ 8(\log_{16}{a})^2 &= 9\log_{16}{a} \\ 8(\log_{16}{a})^2-9\log_{16}{a} + 1 &= 0 \\ 8(\log_{16}{a})^2-8\log_{16}{a} - \log_{16}{a} + 1 &= 0 \\ 8\log_{16}{a}(\log_{16}{a} - 1) - (\log_{16}{a} - 1) &= 0 \\ (\log_{16}{a} - 1)(8\log_{16}{a} - 1) &= 0 \\ \log_{16}{a} = 1 &\textrm{ or } 8\log_{16}{a} = 1 \\ \textrm{Thus, }a &= 16, \sqrt{2} \end{align*} )
that implies either there is a non zero real number c such that,
or
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