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

Instead of converting bases algebraically first, this method assumes a form for $x$ and checks it directly, a quick way to confirm an answer among given choices.

Since every option is written as a power of $2$ or $4$, let $x = 2^{k}$ for some real number $k$, and work out what $k$ must be.

1. The first term becomes $\log_2 x = \log_2 2^{k} = k$, using the basic rule $\log_b b^{k} = k$.
2. For the second term, note $\sqrt{2} = 2^{1/2}$, so $\log_{\sqrt2} x = \log_{2^{1/2}} 2^{k}$. Using the rule $\log_{a^{m}} a^{n} = \dfrac{n}{m}$, this equals $\dfrac{k}{1/2} = 2k$.
3. Adding the two terms: $k + 2k = 3k$, and the equation says this sum equals $48$, so $3k = 48$, giving $k = 16$.

Since $x = 2^{k}$ and $k = 16$, we get $x = 2^{16}$, which is option (A).

Let's summarize:

• Writing $x$ as a power of the smaller base ($2$) turns both logarithms into plain multiples of the exponent $k$, avoiding fraction-heavy change-of-base steps.
• $\log_{\sqrt2} x$ works out to twice $\log_2 x$, not half of it, since $\sqrt2$ is a smaller base than $2$, and a smaller base gives a larger logarithm for the same number.

So $x = 2^{16}$, confirming option (A) as the correct value.