Question

If a positive real \(x\) satisfies the following equation \[ \log_2 x + \log_{\sqrt{2}} x = 48, \] then the value of \(x\) is ____________

• A. \(2^{16}\)
• B. \(4^{16}\)
• C. \(2^{14}\)
• D. \(4^{14}\)

Hint:

Convert the base \(\sqrt{2}\) log into base 2 before combining terms.

Answer 1

The Correct Option is A

Step 1: Substitute $x = 2^y$.
Write $x = 2^y$ so that $\log_2 x = y$ directly.
Since $\sqrt{2} = 2^{1/2}$, the change of base gives $\log_{\sqrt{2}} x = \dfrac{y}{1/2} = 2y$.

Step 2: Solve the resulting equation in y.
The given condition becomes $y + 2y = 48$, so $3y = 48$ and $y = 16$.
Going back to $x = 2^y$ gives $x = 2^{16}$, matching the option written with base 2.

Final Answer:
So $x = 2^{16}$ is the value that satisfies the equation.\[ \boxed{x = 2^{16}} \]