To solve the given logarithmic expression:

$\log_{\sqrt{2}}{12} + \log_{\sqrt{2}}{3} = \log_{\sqrt{2}}{(12 \times 3)} = \log_{\sqrt{2}}{36}$

Step 1: Use the logarithmic product rule:

The product rule of logarithms states:

$\log_b{a} + \log_b{c} = \log_b{(a \times c)}$

Thus:

$\log_{\sqrt{2}}{12} + \log_{\sqrt{2}}{3} = \log_{\sqrt{2}}{(12 \times 3)} = \log_{\sqrt{2}}{36}$

Step 2: Rewrite the logarithmic base:

Since $\sqrt{2} = 2^{1/2}$, the logarithm can be rewritten as:

$\log_{\sqrt{2}}{36} = \frac{\log_2{36}}{\log_2{\sqrt{2}}}$

We know that:

$\log_2{\sqrt{2}} = \log_2{(2^{1/2})} = \frac{1}{2}$

Thus:

$\log_{\sqrt{2}}{36} = \frac{\log_2{36}}{1/2} = 2 \log_2{36}$

Step 3: Simplify $\log_2{36}$:

We can approximate $\log_2{36}$ by recognizing that:

$36 \approx 32 \quad \text{and} \quad \log_2{32} = 5$

So:

$\log_2{36} \approx 5.17$

Step 4: Calculate the final value:

Now we can calculate the approximate value:

$2 \log_2{36} \approx 2 \times 5.17 = 10.34$

Thus, $\log_{\sqrt{2}}{36} \approx 10.34$.

Would you like more details or have any questions?

Here are some related questions to consider:

1. How do you convert logarithms to different bases?
2. What is the change of base formula in logarithms?
3. How do you approximate logarithms for non-exact powers?
4. What are the properties of logarithms involving roots?
5. How can logarithms be applied to exponential equations?

Tip: When working with logarithms involving roots, it's often helpful to express the root as a fractional exponent to simplify calculations.