If $\sqrt[4]{x^3} \cdot \sqrt[3]{x^2} = x^a$ for all values of $x$, what is the value of $a$?

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Simplification is easier when the exponents are written in fractional form:

$$\sqrt[4]{x^3} \cdot \sqrt[3]{x^2} = x^{\frac{3}{4}} \cdot x^{\frac{2}{3}}$$ $$= x^{\frac{3}{4}+\frac{2}{3}}$$ $$= x^{\frac{9}{12}+\frac{8}{12}}$$ $$= x^{\frac{17}{12}}$$