Discussion
If a positive real \(x\) satisfies the following equation
\[ \log_{2} x + \log_{\sqrt{2}} x = 48, \]
then the value of \(x\) is _________________
Answer: A
The value is \(2^{16}\).
Change the second logarithm to base 2. Since \(\sqrt{2} = 2^{1/2}\), \(\log_{\sqrt{2}} x = \frac{\log_{2} x}{1/2} = 2\log_{2} x\). So
\[ \begin{aligned} \log_{2} x + 2\log_{2} x &= 48 \\ 3\log_{2} x &= 48 \\ \log_{2} x &= 16,\quad x = 2^{16} \end{aligned} \]
\(4^{16} = 2^{32}\) doubles the exponent, and \(2^{14}\) and \(4^{14} = 2^{28}\) come from mishandling the base \(\sqrt{2}\), for example treating \(\log_{\sqrt{2}}\) as a factor other than 2.
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