1. **State the problem:** Simplify the expression $$a^{\log_{a^2} b^4 + \log_a (7b^2)}$$ where $$a, b > 0$$ and $$a \neq 1$$. 2. **Recall logarithm change of base and properties:** - $$\log_{a^2} b^4 = \frac{\log_a b^4}{\log_a a^2}$$ - $$\log_a b^4 = 4 \log_a b$$ - $$\log_a a^2 = 2$$ - Also, $$\log_a (7b^2) = \log_a 7 + \log_a b^2 = \log_a 7 + 2 \log_a b$$ 3. **Rewrite the exponent:** $$\log_{a^2} b^4 + \log_a (7b^2) = \frac{4 \log_a b}{2} + \log_a 7 + 2 \log_a b = 2 \log_a b + \log_a 7 + 2 \log_a b$$ 4. **Combine like terms:** $$2 \log_a b + 2 \log_a b + \log_a 7 = 4 \log_a b + \log_a 7$$ 5. **Use logarithm addition property:** $$4 \log_a b + \log_a 7 = \log_a b^4 + \log_a 7 = \log_a (7 b^4)$$ 6. **Rewrite the original expression:** $$a^{\log_a (7 b^4)}$$ 7. **Use the property $$a^{\log_a x} = x$$:** $$a^{\log_a (7 b^4)} = 7 b^4$$ 8. **Check if this matches any options:** Options are: b/a, b, none of them, 7, a^7. Since $$7 b^4$$ is not among the options, the answer is **none of them**. **Final answer:** none of them