log question (1 Viewer)

Mahan1 · Feb 19, 2017

If
$log_{a}16 + log_{\sqrt{2}}a = 9$
then find a.

1729 · Feb 19, 2017

Mahan1 said:
If
$log_{a}16 + log_{\sqrt{2}}a = 9$
then find a.

$$\noindent There may be a shorter method but this is what I have so far; \\ (choosing base $\sqrt{2}$ also works) \\ \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*}$

Mahan1 · Feb 20, 2017

1729 said:
$$\noindent There may be a shorter method but this is what I have so far; \\ (choosing base $\sqrt{2}$ also works) \\ \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*}$

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:

$log_{a}\ 16 +log_{\sqrt{2}}\ a = 9 \ \ \ \ \ \ (1)$

$log_{a}\ 16 - 8 = -(log_{\sqrt{2}}\ a - 1)$

$log_{a}\ \frac{16}{a^{8}} = -(log_{\sqrt{2}}\ \frac{a}{\sqrt{2}}) \ \textrm{(By log's rule)}$

$log_{a}\ (\frac{\sqrt{2}}{a})^{8} = (log_{\sqrt{2}}\ \frac{\sqrt{2}}{a})$

that implies either there is a non zero real number c such that,

$a^c = (\sqrt{2})^{8c} \Rightarrow a = 16$

or

$\frac{\sqrt{2}}{a} = 1 \Rightarrow a =\sqrt{2} \ \textrm{(when c =0)}$

Drongoski · Feb 20, 2017

Very interesting question. I used:

$\frac {log _{\sqrt 2} 16}{log_{\sqrt 2} a} + log_{\sqrt 2} a = 9 \implies (log_{\sqrt 2} a)^2 - 9log_{\sqrt 2} a + 8 = 0 \implies \\ \\ m^2-9m+8=0\\ \\ \therefore (m-1)(m-8) = 0 \implies m = 1 $ or $ 8\\ \\m=1: $ $ log_{\sqrt 2} a = 1 \implies a = (\sqrt 2)^1 = \sqrt 2 \\ \\ m = 8: $ $ log_{\sqrt 2} a = 8 \implies a = (\sqrt 2)^8 = 2^4 = 16$

Drongoski · Feb 21, 2017

Which version would you prefer?

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log question (1 Viewer)

Mahan1

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