Logarithmic Equation and Maximum Integer

Using base-change and power rules for logarithms: $\log_8(a+b) = \frac{1}{3} \log_2(a+b) \implies \frac{\log_8(a+b)}{\log_2 c} = \frac{1}{3} \log_c(a+b)$ $\log_{27}(a-b) = \frac{1}{3} \log_3(a-b) \implies \frac{\log_{27}(a-b)}{\log_3 c} = \frac{1}{3} \log_c(a-b)$

Adding these two terms: $\frac{1}{3} \left[ \log_c(a+b) + \log_c(a-b) \right] = \frac{2}{3}$ $\log_c((a+b)(a-b)) = 2 \implies a^2 - b^2 = c^2 \implies a^2 = b^2 + c^2$

We are given $10 \ge b \ge c > 0$. To maximize $a$, we maximize $b^2 + c^2$: $b \le 10 \quad \text{and} \quad c \le 10$ $a^2 = b^2 + c^2 \le 10^2 + 10^2 = 200$ $a \le \sqrt{200} \approx 14.14$

Thus, the greatest integer value of $a$ is 14.