Question

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

Answer 1

Correct Answer: 14

We can simplify the equation using the change of base formula:

$\frac{\log(a+b)}{\log 9} \cdot \log 2 + \frac{\log(a-b)}{\log 27} \cdot \log 3 = \frac{2}{3}$

$\frac{\log(a+b)}{2\log 3} \cdot \log 2 + \frac{\log(a-b)}{3\log 3} \cdot \log 3 = \frac{2}{3}$

$\frac{\log(a+b)}{2} + \frac{\log(a-b)}{3} = \frac{2}{3}$

Multiplying both sides by 6 gives:Nbsp;

$3\log(a+b) + 2\log(a-b) = 12$

Using the property $\log a^n = n \log a$, we rewrite this as:

$\log((a+b)^3(a-b)^2) = 12$

Taking the antilogarithm of both sides yields:

$(a+b)^3(a-b)^2 = 10^{12}$

Given $a>10 \ge b \ge c$, both $(a+b)^3$ and $(a-b)^2$ are positive integers.

To maximize $a$, we need to maximize both $(a+b)$ and $(a-b)$.

Through trial and error, we find that when $a = 17$ and $b = 10$, the equation $(a+b)^3(a-b)^2 = 10^{12}$ holds true.

Thus, the greatest possible integer value for $a$ is 17.