Solve the logarithmic equation: log9 4 + log3 3^x = 3

Using the logarithm property $\text{log}_b b^p = p$:

$$\text{log}_3 3^x = x$$

The equation now becomes:

$$\text{log}_9 4 + x = 3$$

Subtract $\text{log}_9 4$ from both sides:

$$x = 3 - \text{log}_9 4$$

We can change the base of the logarithm. Since $9 = 3^2$, we can rewrite $\text{log}_9 4$ using base 3.
Using the property $\text{log}_{b^k} A = \frac{1}{k} \text{log}_b A$:

$$\text{log}_9 4 = \text{log}_{3^2} 4 = \frac{1}{2} \text{log}_3 4$$

Now, use the property $p \text{log}_b A = \text{log}_b A^p$:

$$\frac{1}{2} \text{log}_3 4 = \text{log}_3 4^{1/2} = \text{log}_3 \sqrt{4} = \text{log}_3 2$$

$$x = 3 - \text{log}_3 2$$

The exact solution for $x$ is: