Instead of solving the equation symbolically, we can test each answer choice directly in the given equation \( \log_2 x + \log_{\sqrt{2}} x = 48 \) and see which one balances. First it helps to note the general rule \( \log_{\sqrt{2}} x = 2\log_2 x \), because raising the base to the power \(1/2\) doubles the log value.
- $2^{16}$: Here $\log_2 x = 16$, so $\log_{\sqrt{2}} x = 2(16) = 32$. Sum $= 16 + 32 = 48$. This matches the right-hand side exactly.
- $4^{16} = 2^{32}$: Here $\log_2 x = 32$, so $\log_{\sqrt{2}} x = 64$. Sum $= 32 + 64 = 96$, not 48.
- $2^{14}$: Here $\log_2 x = 14$, so $\log_{\sqrt{2}} x = 28$. Sum $= 14 + 28 = 42$, not 48.
- $4^{14} = 2^{28}$: Here $\log_2 x = 28$, so $\log_{\sqrt{2}} x = 56$. Sum $= 28 + 56 = 84$, not 48.
Only $x = 2^{16}$ makes both sides of the equation equal, so it is the value that satisfies the condition given in the question.
Let's sum up:
- Changing the base from $\sqrt{2}$ to $2$ doubles the log value, since $\sqrt{2}$ is the square root, that is, the one-half power, of $2$.
- Testing each option this way avoids solving the equation from scratch and confirms the same answer.
So $x = 2^{16}$, which is option (A).