1. **State the problem:** We want to simplify the expression $$\log \left( \sqrt{\frac{7^2 t^3 p}{d^6 b^2}} \right)$$. 2. **Rewrite the square root as an exponent:** Recall that $$\sqrt{x} = x^{1/2}$$, so we have: $$\log \left( \frac{7^2 t^3 p}{d^6 b^2} \right)^{1/2}$$ 3. **Apply the logarithm power rule:** $$\log (x^a) = a \log x$$, this gives: $$\frac{1}{2} \log \left( \frac{7^2 t^3 p}{d^6 b^2} \right)$$ 4. **Use the logarithm quotient rule:** $$\log \left( \frac{A}{B} \right) = \log A - \log B$$, so: $$\frac{1}{2} \left( \log(7^2 t^3 p) - \log(d^6 b^2) \right)$$ 5. **Use the logarithm product rule:** $$\log (xyz) = \log x + \log y + \log z$$, thus: $$\frac{1}{2} \left( \log 7^2 + \log t^3 + \log p - \log d^6 - \log b^2 \right)$$ 6. **Apply the log power rule on each term:** $$\log 7^2 = 2 \log 7$$ $$\log t^3 = 3 \log t$$ $$\log d^6 = 6 \log d$$ $$\log b^2 = 2 \log b$$ 7. **Substitute back:** $$\frac{1}{2} \left( 2 \log 7 + 3 \log t + \log p - 6 \log d - 2 \log b \right)$$ 8. **Distribute the $$\frac{1}{2}$$:** $$ = \log 7 + \frac{3}{2} \log t + \frac{1}{2} \log p - 3 \log d - \log b$$ **Final answer:** $$\boxed{\log 7 + \frac{3}{2} \log t + \frac{1}{2} \log p - 3 \log d - \log b}$$