The given equation is:

\[
\log_a 2 + \log_a 1 + \log_a \sqrt{a} = 3 \cdot 2 \log_a a
\]

### Step 1: Simplify each term on the left-hand side
- \(\log_a 2\) remains as it is.
- \(\log_a 1 = 0\) (since \(\log_a 1 = 0\) for any base \(a\)).
- \(\log_a \sqrt{a} = \log_a a^{1/2} = \frac{1}{2} \log_a a = \frac{1}{2}\) (because \(\log_a a = 1\)).

Thus, the left-hand side becomes:
\[
\log_a 2 + 0 + \frac{1}{2} = \log_a 2 + \frac{1}{2}
\]

### Step 2: Simplify the right-hand side
The right-hand side is:
\[
3 \cdot 2 \log_a a = 3 \cdot 2 \cdot 1 = 6
\]
(since \(\log_a a = 1\)).

### Step 3: Set up the simplified equation
Now we have:
\[
\log_a 2 + \frac{1}{2} = 6
\]

### Step 4: Solve for \(\log_a 2\)
Subtract \(\frac{1}{2}\) from both sides:
\[
\log_a 2 = 6 - \frac{1}{2} = \frac{12}{2} - \frac{1}{2} = \frac{11}{2}
\]

### Step 5: Exponentiate to solve for \(a\)
We know that:
\[
\log_a 2 = \frac{11}{2}
\]
This implies:
\[
a^{\frac{11}{2}} = 2
\]

Now raise both sides to the power of \(\frac{2}{11}\):
\[
a = 2^{\frac{2}{11}}
\]

Thus, the value of \(a\) is \(2^{\frac{2}{11}}\).

Would you like further details or have any other questions?

Here are 5 follow-up questions based on this explanation:
1. What is the general formula for solving logarithmic equations?
2. How would the equation change if the base of the logarithm was different?
3. Can you explain the relationship between exponents and logarithms in more depth?
4. How can we interpret the result \(2^{\frac{2}{11}}\) in decimal form?
5. How would you apply these logarithmic techniques to real-world problems?

**Tip:** Remember that logarithmic and exponential functions are inverses of each other, which is key in solving equations like this one!

log_a 2 + log_a 1 + log_a √a = 3 * 2 * log_a a

This solution demonstrates how to solve the logarithmic equation log_a 2 + log_a √a = 6. Key concepts include properties of logarithms and exponentiation. Suitable for high school students, Grades 9-12.

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