If a, b and c are positive real numbers such that $a > 10 \ge b \ge c$ and $\frac{\log_8(a+b)}{\log_2 c} + \frac{\log_{27}(a - b)}{\log_3 c} = \frac{2}{3}$, then the greatest possible integer value of a is

Given Information:

• $a$, $b$, and $c$ are positive real numbers
• $a > 10 \geq b \geq c$
• $\frac{\log_8(a+b)}{\log_2 c} + \frac{\log_{27}(a - b)}{\log_3 c} = \frac{2}{3}$

---

The bases 8 and 27 look intimidating, but here's the key insight: both are perfect powers of simpler numbers.

Since $8 = 2^3$ and $27 = 3^3$, we can use the property:

$\log_{a^n}(x) = \frac{1}{n}\log_a(x)$

This gives us:

• $\log_8(a+b) = \log_{2^3}(a+b) = \frac{1}{3}\log_2(a+b)$
• $\log_{27}(a-b) = \log_{3^3}(a-b) = \frac{1}{3}\log_3(a-b)$

When the base is a power, we can "bring down" the exponent as a fraction in front.

---

Our equation becomes:

$\frac{\tfrac{1}{3}\log_2(a+b)}{\log_2 c} + \frac{\tfrac{1}{3}\log_3(a-b)}{\log_3 c} = \frac{2}{3}$

Factor out $\frac{1}{3}$:

$\frac{1}{3}\left(\frac{\log_2(a+b)}{\log_2 c} + \frac{\log_3(a-b)}{\log_3 c}\right) = \frac{2}{3}$

$\frac{\log_2(a+b)}{\log_2 c} + \frac{\log_3(a-b)}{\log_3 c} = 2$

---

Here's a powerful property: $\frac{\log_a(x)}{\log_a(y)} = \log_y(x)$

This comes from the change of base formula. When we have the same base in numerator and denominator, we can "flip" the denominator to become the new base.

Applying this:

• $\frac{\log_2(a+b)}{\log_2 c} = \log_c(a+b)$
• $\frac{\log_3(a-b)}{\log_3 c} = \log_c(a-b)$

Our equation becomes:

$\log_c(a+b) + \log_c(a-b) = 2$

---

Using the property $\log_a(x) + \log_a(y) = \log_a(xy)$:

$\log_c((a+b)(a-b)) = 2$

---

$(a+b)(a-b) = a^2 - b^2$ (difference of squares)

So: $\log_c(a^2 - b^2) = 2$

---

If $\log_c(x) = 2$, then by definition: $c^2 = x$

Therefore: $c^2 = a^2 - b^2$

$a^2 = b^2 + c^2$

This is our key relationship!

---

To find the greatest possible integer value of $a$, we need to maximize $a^2 = b^2 + c^2$.

From our constraints $a > 10 \geq b \geq c$:

• To maximize $b^2$, we choose $b = 10$ (the maximum allowed)
• To maximize $c^2$, we choose $c = 10$ (since $b \geq c$ allows $c = b = 10$)

Therefore:

$a^2 = 10^2 + 10^2 = 100 + 100 = 200$

$a = \sqrt{200} = \sqrt{100 \times 2} = 10\sqrt{2} \approx 14.14$

---

Let's check: $a = 10\sqrt{2} \approx 14.14 > 10$

Since we need the greatest possible integer value, and $a \approx 14.14$, the answer is 14.

Key Insight: This problem transforms a complex logarithmic equation into a simple Pythagorean-like relationship $a^2 = b^2 + c^2$, then uses constraint optimization to find the maximum value.