1. **State the problem:** Simplify the expression $$\frac{(\log_3 567 - \log_3 7) + 10^{\log 6}}{7^{1+10} \cdot 7^{3 - \log_{11} \sqrt{x^{x}}}}$$. 2. **Recall logarithm and exponent rules:** - $\log_a b - \log_a c = \log_a \frac{b}{c}$. - $a^m \cdot a^n = a^{m+n}$. - $10^{\log 6} = 6$ because $10^{\log 6} = 6$ (log base 10). - $\sqrt{x^x} = (x^x)^{\frac{1}{2}} = x^{\frac{x}{2}}$. - $\log_a b^c = c \log_a b$. 3. **Simplify numerator:** $$\log_3 567 - \log_3 7 = \log_3 \frac{567}{7} = \log_3 81$$ Since $81 = 3^4$, $$\log_3 81 = 4$$ Also, $$10^{\log 6} = 6$$ So numerator becomes: $$4 + 6 = 10$$ 4. **Simplify denominator:** Combine powers of 7: $$7^{1+10} \cdot 7^{3 - \log_{11} \sqrt{x^{x}}} = 7^{11} \cdot 7^{3 - \log_{11} x^{\frac{x}{2}}} = 7^{11 + 3 - \log_{11} x^{\frac{x}{2}}} = 7^{14 - \log_{11} x^{\frac{x}{2}}}$$ Use logarithm power rule: $$\log_{11} x^{\frac{x}{2}} = \frac{x}{2} \log_{11} x$$ So denominator is: $$7^{14 - \frac{x}{2} \log_{11} x}$$ 5. **Final simplified expression:** $$\frac{10}{7^{14 - \frac{x}{2} \log_{11} x}}$$ This is the simplified form of the original expression.