CAT 2024 Slot 2 QA Question 20

The question says: If a, b, and c are positive real numbers such that a > 10 ≥ b ≥ c, and the equation (log base 8 of (a + b)) divided by (log base 2 of c) plus (log base 27 of (a - b)) divided by (log base 3 of c) equals 2/3. We need to find the greatest possible integer value of a.

Hmm, okay. So, variables a, b, c are positive, with a being the largest, greater than 10, and b and c are less than or equal to 10, with b being at least as big as c.

The equation involves logarithms with different bases. Let me write it down to visualize better:

(log₈(a + b)) / (log₂ c) + (log₂₇(a - b)) / (log₃ c) = 2/3.

logarithms can be converted from one base to another using the change of base formula. Maybe that can help here. The change of base formula is log_b x = (log_k x) / (log_k b), right? So, if I can express all logs with the same base, maybe things will simplify.

Looking at the denominators, we have log₂ c and log₃ c. if I can express the numerators in terms of logs with base 2 or 3.

First term: log₈(a + b). Since 8 is 2³, log₈ x is equal to (log₂ x) / (log₂ 8) which is (log₂ x) / 3. So, log₈(a + b) = (log₂(a + b)) / 3.

Similarly, the second term: log₂₇(a - b). Since 27 is 3³, log₂₇ x = (log₃ x) / 3. So, log₂₇(a - b) = (log₃(a - b)) / 3.

So, substituting these into the original equation:

[(log₂(a + b)/3) / (log₂ c)] + [(log₃(a - b)/3) / (log₃ c)] = 2/3.

Simplify each term:

First term: (log₂(a + b)/3) / (log₂ c) = (log₂(a + b)) / (3 log₂ c).

Second term: (log₃(a - b)/3) / (log₃ c) = (log₃(a - b)) / (3 log₃ c).

So, the equation becomes:

[log₂(a + b) / (3 log₂ c)] + [log₃(a - b) / (3 log₃ c)] = 2/3.

Factoring out 1/3 from both terms:

(1/3) [log₂(a + b)/log₂ c + log₃(a - b)/log₃ c] = 2/3.

Multiply both sides by 3 to eliminate the 1/3:

log₂(a + b)/log₂ c + log₃(a - b)/log₃ c = 2.

Now, Note that log_b x / log_b y is equal to log_y x. Because log_b x = (ln x)/(ln b) and log_b y = (ln y)/(ln b), so their ratio is (ln x)/(ln y) = log_y x.

So, applying that here:

log₂(a + b)/log₂ c = log_c(a + b).

Similarly, log₃(a - b)/log₃ c = log_c(a - b).

So, the equation simplifies to:

log_c(a + b) + log_c(a - b) = 2.

Using logarithm properties, log_c(A) + log_c(B) = log_c(A * B). So:

log_c[(a + b)(a - b)] = 2.

Which means:

(a + b)(a - b) = c².

Because log_c(X) = Y implies X = c^Y.

So, (a + b)(a - b) = c².

Simplify the left side: it's a difference of squares, so a² - b² = c².

Therefore, a² = b² + c².

Hmm, interesting. So, a, b, c satisfy the Pythagorean theorem. That's a key insight.

Given that a, b, c are positive real numbers, and a > 10 ≥ b ≥ c, so a is the hypotenuse of a right-angled triangle with sides b and c.

But wait, in a right-angled triangle, the hypotenuse is the longest side, which is consistent with a > b and a > c.

But in our case, a > 10, and b and c are ≤10. So, a must be greater than 10, and b and c are at most 10.

But since a² = b² + c², and b and c are ≤10, the maximum possible a would be when b and c are as large as possible, i.e., 10 each. So, a² = 10² + 10² = 200, so a = \ ≈14.142. So, the greatest integer less than or equal to 14.142 is 14.

But wait, the question says "greatest possible integer value of a". So, 14 is the maximum integer less than \. But we need to confirm if a can actually reach 14, or if there are constraints that might make 14 impossible.

Wait, let's think again. The problem states that a >10, and b and c are ≤10, with b ≥ c. So, to maximize a, we need to maximize b and c. The maximum b and c can be is 10 each. So, a² = 10² +10²=200, so a= \$\sqrt{200}$ = 10*\$\sqrt{2}$$≈14.142. So, the maximum integer less than this is 14. But can a be exactly 14? Let's check.

If a=14, then a²=196. So, 196 = b² + c². Since b and c are ≤10, the maximum b² + c² can be is 100 + 100=200. So, 196 is less than 200, which is possible. So, we can have b and c such that b² + c²=196.

But wait, we also have the original equation involving logs. So, we need to ensure