Question:

Step 1: Evaluate the first expression. $$7^{-\log_7 6} + 81^{(1-\log_9 2)}$$ Using the property $a^{\log_a x} = x$, we have $7^{-\log_7 6} = 7^{\log_7 (6^{-1})} = 6^{-1} = \frac{1}{6}$. For the second term, we rewrite $81$ as $9^2$: $$81^{(1-\log_9 2)} = (9^2)^{(1-\log_9 2)} = 9^{2(1-\log_9 2)} = 9^{2 - 2\log_9 2} = 9^2 \cdot 9^{-2\log_9 2}$$ $$= 81 \cdot (9^{\log_9 2})^{-2} = 81 \cdot (2)^{-2} = 81 \cdot \frac{1}{2^2} = \frac{81}{4}$$ Adding the two terms: $$\frac{1}{6} + \frac{81}{4} = \frac{2}{12} + \frac{243}{12} = \frac{245}{12}$$ Since $\frac{245}{12}$ is not an integer, the first expression is not an integer. Step 2: Evaluate the second expression. $$\log_6 3 \cdot \log_6 12 + (\log_6 2)^2$$ We use the logarithm properties $\log_b (xy) = \log_b x + \log_b y$ and $\log_b (x^k) = k \log_b x$. Let $x = \log_6 2$ and $y = \log_6 3$. Then $\log_6 12 = \log_6 (2^2 \cdot 3) = 2\log_6 2 + \log_6 3 = 2x+y$. The expression becomes: $$y(2x+y) + x^2 = 2xy + y^2 + x^2 = (x+y)^2$$ Since $x+y = \log_6 2 + \log_6 3 = \log_6 (2 \cdot 3) = \log_6 6 = 1$. Therefore, the expression is: $$(1)^2 = 1$$ Since $1$ is an integer, the second expression is an integer. Step 3: Evaluate the third expression. $$\frac{1}{\log_5 3} + \frac{1}{\log_6 3} - \frac{1}{\log_{10} 3}$$ Using the change of base formula $\frac{1}{\log_b a} = \log_a b$: $$= \log_3 5 + \log_3 6 - \log_3 10$$ Using the logarithm properties $\log_b x + \log_b y = \log_b (xy)$ and $\log_b x - \log_b y = \log_b \left(\frac{x}{y}\right)$: $$= \log_3 \left(\frac{5 \cdot 6}{10}\right) = \log_3 \left(\frac{30}{10}\right) = \log_3 3$$ Since $\log_3 3 = 1$, the expression is: $$1$$ Since $1$ is an integer, the third expression is an integer. Step 4: Evaluate the fourth expression. $$(\sqrt[3]{2} + \sqrt[3]{5})(\sqrt[3]{4} - \sqrt[3]{10} + \sqrt[3]{25})$$ This expression matches the algebraic identity for the sum of cubes: $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$. Let $a = \sqrt[3]{2}$ and $b = \sqrt[3]{5}$. Then $a^2 = (\sqrt[3]{2})^2 = \sqrt[3]{4}$. $ab = \sqrt[3]{2} \cdot \sqrt[3]{5} = \sqrt[3]{10}$. $b^2 = (\sqrt[3]{5})^2 = \sqrt[3]{25}$. Substituting these into the identity: $$(\sqrt[3]{2})^3 + (\sqrt[3]{5})^3 = 2 + 5 = 7$$ Since $7$ is an integer, the fourth expression is an integer.