To solve $\log_9 3$, we want to determine the exponent $x$ such that:

$9^x = 3$

We can express 9 as $3^2$. Therefore, the equation becomes:

$(3^2)^x = 3$

Simplifying, we have:

$3^{2x} = 3^1$

Since the bases are the same, we can equate the exponents:

$2x = 1$

Solving for $x$:

$x = \frac{1}{2}$

Thus, $\log_9 3 = \frac{1}{2}$. The correct answer is $\frac{1}{2}$.

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Related Questions:

1. How do you solve logarithms with different bases?
2. What is the change of base formula in logarithms?
3. How do you evaluate $\log_4 2$?
4. What is the relationship between logarithms and exponents?
5. How do you simplify expressions involving logarithms?
6. What is the value of $\log_9 81$?
7. How do you solve logarithmic equations with multiple logarithms?
8. What are common logarithms and how are they used?

Tip: Remember that logarithms can be converted to a different base using the change of base formula: $\log_b a = \frac{\log_c a}{\log_c b}$.