Solving the Logarithm Equation $\log_2 x + \log_{\sqrt{2}} x = 48$

Key Logarithm Properties Used

• Change of base property: $\log_{b^k} a = \frac{1}{k} \log_b a$.
• Combining like terms.

Step-by-Step Solution

1. Simplify the second term: The base of the second logarithm is $\sqrt{2}$, which can be written as $2^{1/2}$. Using the property $\log_{b^k} a = \frac{1}{k} \log_b a$, we have:

   $\log_{\sqrt{2}} x = \log_{2^{1/2}} x = \frac{1}{1/2} \log_2 x = 2 \log_2 x$.

2. Substitute back into the equation: Replace $\log_{\sqrt{2}} x$ with $2 \log_2 x$ in the original equation:

   $\log_2 x + 2 \log_2 x = 48$.

3. Combine logarithmic terms: Add the terms involving $\log_2 x$:

   $3 \log_2 x = 48$.

4. Isolate the logarithm: Divide both sides by 3:

   $\log_2 x = \frac{48}{3}$

   $\log_2 x = 16$.

5. Solve for $x$: Convert the logarithmic equation to its exponential form. If $\log_b y = c$, then $y = b^c$.

   $x = 2^{16}$.

Final Answer

The value of $x$ that satisfies the equation $\log_2 x + \log_{\sqrt{2}} x = 48$ is $2^{16}$.