How to approach this tricky log problem

Question

Consider the following equation:

$$(7 + 4\sqrt{3})^{t^2 - 5t + 5} + (7 - 4\sqrt{3})^{t^2 - 5t + 5} = 14$$

I have tried and exhausted all the methods I know of and resorted to brute force to solve this problem and got $t = 4$, $t = 1$. However I was wondering if there was some sort of elegant solution to prove the answer algebraically.

Answer 1

Your problem is just the equivalent of

$$(7 + 4\sqrt{3})^{p(t)} + (7 - 4\sqrt{3})^{p(t)} = 14$$

Where $p(t) = t^2 - 5t + 5$. Notice that if $p(t) = 1$ you're done since

$$(7 + 4\sqrt{3})^1 + (7 - 4\sqrt{3})^1 = 14$$

So: what are the values in $\mathbb{R}$ for which $p(t) = 1$?

(Notice that also $p(t) = -1$ works... Can you see it?)

Answer 2

Setting $a(t)=(7+4\sqrt 3)^{t^2-5t+5}$ so thar $\frac 1{a(t)}=(7-4\sqrt 3)^{t^2-5t+5}$, we get $$a(t)+\frac 1{a(t)}=14$$ and hence $a(t)=7+4\sqrt 3$ or $a(t)=7-4\sqrt 3$. So, we have $$t^2-5t+5=1 \text{ or}\;(-1)$$ Now, solve for $t$.