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:

Use \( \log_{\sqrt{2}} x = 2\log_2 x \) since \(\sqrt{2}\) is the one-half power of 2.

Answer 1

The Correct Option is A

Concept:
To combine logarithms with different bases, first convert every term to the same base using the change of base formula \( \log_b a = \dfrac{\log_c a}{\log_c b} \).

Step 1: Rewrite \(\log_{\sqrt{2}} x\) in 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 \[ \log_{\sqrt{2}} x = \frac{\log_2 x}{1/2} = 2\log_2 x. \]

Step 2: Substitute back into the given equation.
\[ \log_2 x + 2\log_2 x = 48 \]
\[ 3\log_2 x = 48 \]

Step 3: Solve for \(\log_2 x\), then for \(x\).
\[ \log_2 x = \frac{48}{3} = 16 \]
By the definition of logarithm, \(\log_2 x = 16\) means \(x = 2^{16}\).

Step 4: Check the other options are not equal to \(2^{16}\).
\(4^{16} = (2^2)^{16} = 2^{32}\), far larger than \(2^{16}\), so option (B) is wrong.
\(2^{14}\) and \(4^{14}=2^{28}\) both give a different exponent of 2 than 16, so options (C) and (D) fail the equation too.

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