1. **Stating the problem:** Given the equations $\log_{\sqrt{a}} b = 4$ and $b^{\log_{\sqrt{b}} 3} = a$, find the value of $b$. 2. **Rewrite the logarithms in terms of base 10 or natural logs:** Recall that $\log_{x} y = \frac{\log y}{\log x}$. From the first equation: $$\log_{\sqrt{a}} b = 4 \implies \frac{\log b}{\log \sqrt{a}} = 4$$ Since $\sqrt{a} = a^{1/2}$, then $\log \sqrt{a} = \frac{1}{2} \log a$. So, $$\frac{\log b}{\frac{1}{2} \log a} = 4 \implies \frac{\log b}{\log a / 2} = 4$$ 3. **Simplify the above:** $$\frac{\log b}{\log a / 2} = 4 \implies \log b = 4 \times \frac{\log a}{2} = 2 \log a$$ 4. **Rewrite the second equation:** $$b^{\log_{\sqrt{b}} 3} = a$$ Again, rewrite the logarithm: $$\log_{\sqrt{b}} 3 = \frac{\log 3}{\log \sqrt{b}} = \frac{\log 3}{\frac{1}{2} \log b} = \frac{2 \log 3}{\log b}$$ So, $$b^{\frac{2 \log 3}{\log b}} = a$$ 5. **Simplify the exponentiation:** Recall that $b^{\frac{\log k}{\log b}} = k$ for any positive $k$. Here, $$b^{\frac{2 \log 3}{\log b}} = \left(b^{\frac{\log 3}{\log b}}\right)^2 = 3^2 = 9$$ So, $$a = 9$$ 6. **Use the relation from step 3:** $$\log b = 2 \log a = 2 \log 9$$ Since $\log 9 = \log 3^2 = 2 \log 3$, then $$\log b = 2 \times 2 \log 3 = 4 \log 3$$ 7. **Rewrite $\log b$:** $$\log b = 4 \log 3 = \log 3^4 = \log 81$$ 8. **Conclude the value of $b$:** Since $\log b = \log 81$, then $$b = 81$$ **Final answer:** $$\boxed{81}$$