AIME Logarithm problem!

This problem is not based around solving individual $a$ , $b$ , and $c$ . Instead, it’s more about manipulating the values given to get the desired outcome.

Let’s look at the first value in the final outcome we’re supposed to find: $a^{(\log_3 7)^2}$ . Perhaps if we notice something with the first value, we can apply it to the others since they are all relatively similar.

Substituting in $a^{(\log_3 7)} = 27$ , we get $27^{\log_3 7}$ . Recall that if we have $a^{\log_a b}$ , it is equivalent to $b$ . Here, we can express $27$ as $3^3$ , so the resulting expression is $3^{3\log_3 7}$ , or $3^{\log_3 343}$ , which simplifies to $343$ . Aha! Now we have found a method that we can apply similarly to all of the other values.

Applying this method to $b$ , we get that $b^{(\log_7 11)^2}$ is equivalent to $49^{\log_7 11}$ which simplifies to $7^{2\log_7 11}$ , where we end up with $121$ .

$c$ is a similar process as well, where $c^{(\log_{11} 25)^2}$ simplifies down to $\sqrt{11}^{\log_{11} 25}$ , which is $11^{(\frac{1}{2}\log_{11} 25)}$ , which simplifies to $5$

$343 + 121 + 5$ gives us our final answer of $\boxed{469}$