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 \(\log_{\sqrt{2}} x\) to base 2 using the change of base rule before adding the two terms.

Answer 1

The Correct Option is A

Step 1: Write both terms in the same base.
The equation has two logarithms with different bases, base \(2\) and base \(\sqrt{2}\). To add them we first convert \(\log_{\sqrt{2}} x\) into base \(2\) using the change of base rule \(\log_b a = \dfrac{\log_c a}{\log_c b}\).
\[ \log_{\sqrt{2}} x = \frac{\log_2 x}{\log_2 \sqrt{2}} \]

Step 2: Simplify \(\log_2 \sqrt{2}\).
Since \(\sqrt{2} = 2^{1/2}\), we get \(\log_2 \sqrt{2} = \log_2 2^{1/2} = \frac{1}{2}\).
So
\[ \log_{\sqrt{2}} x = \frac{\log_2 x}{1/2} = 2\log_2 x \]

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

Step 4: Solve for \(\log_2 x\), then for \(x\).
Dividing both sides by \(3\):
\[ \log_2 x = 16 \]
By the definition of a logarithm, \(\log_2 x = 16\) means \(x = 2^{16}\).

Step 5: Check why the other options are wrong.
Option (B) \(4^{16} = 2^{32}\) and option (D) \(4^{14} = 2^{28}\) both come from wrongly treating \(\log_{\sqrt2} x\) as \(\tfrac{1}{2}\log_2 x\) instead of \(2\log_2 x\), which flips the change of base fraction upside down. Option (C) \(2^{14}\) would only be right if the sum in the equation were \(42\) instead of \(48\).

Step 6: Final Answer.
\[ \boxed{x = 2^{16}} \]