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 log base sqrt(2) to base 2 using the change-of-base rule before combining terms.

Answer 1

The Correct Option is A

Step 1: Understanding the Question.
We are given the equation \( \log_2 x + \log_{\sqrt{2}} x = 48 \) and need to find the positive real value of \(x\) that satisfies it.

Step 2: Key Formula or Approach.
Use the change-of-base rule \( \log_b a = \dfrac{\log_c a}{\log_c b} \) to write \( \log_{\sqrt{2}} x \) in terms of \( \log_2 x \).

Step 3: Detailed Explanation.
Convert \( \log_{\sqrt{2}} x \) to base 2:
\[ \log_{\sqrt{2}} x = \frac{\log_2 x}{\log_2 \sqrt{2}} \] Since \( \sqrt{2} = 2^{1/2} \), we get \( \log_2 \sqrt{2} = \frac{1}{2} \). So
\[ \log_{\sqrt{2}} x = \frac{\log_2 x}{1/2} = 2\log_2 x \] Substitute this back into the original equation:
\[ \log_2 x + 2\log_2 x = 48 \] \[ 3\log_2 x = 48 \] \[ \log_2 x = 16 \] Converting back from log form:
\[ x = 2^{16} \]
Step 4: Why the other options are wrong.
Option (B) \(4^{16} = 2^{32}\) and option (D) \(4^{14} = 2^{28}\) both come from mis-simplifying \( \log_{\sqrt{2}} x \) (for example, mistakenly treating it as \( \frac{1}{2}\log_2 x\) instead of \( 2\log_2 x\), or making an arithmetic slip while solving for the exponent). Option (C) \(2^{14}\) comes from an arithmetic error in dividing 48 by 3. None of these match the correct working above.

Step 5: Final Answer.
The value of \(x\) is \(2^{16}\). \[ \boxed{x = 2^{16}} \]