4. Given that $\alpha$ is an acute angle, and $3 \sin \alpha+4 \cos \alpha=5$. Then $\tan \alpha=$ $\qquad$ .
4. $\frac{3}{4}$.

In $\triangle ABC$, $\angle ACB=90^{\circ}$, let $\angle A=\alpha$. Then $\sin \alpha=\frac{a}{c}, \cos \alpha=\frac{b}{c}$. From the given information and the Pythagorean theorem, we have
$$
\left\{\begin{array}{l}
\frac{3 a}{c}+\frac{4 b}{c}=5, \\
a^{2}+b^{2}=c^{2} .
\end{array}\right.
$$

Solving these equations, we get $a=\frac{3 c}{5}, b=\frac{4 c}{5}$. Therefore, $\tan \alpha=\frac{a}{b}=\frac{3}{4}$.