We square both sides of the equation to get rid of the square roots:
$2\sqrt{t-2} = \sqrt[4]{(7 - t)^2}$
Squaring both sides again, we get:
$4(t-2) = (7 - t)^2$
Expanding the right side, we have:
$4t - 8 = 49 - 14t + t^2$
Rearranging the terms, we have a quadratic equation:
$t^2 - 18t + 57 = 0$
Factoring the quadratic, we have:
$(t - 3)(t - 19) = 0$
Therefore, the possible values for $t$ are 3 and 19.
However, we need to check if these values satisfy the original equation.
Checking $t = 3$, we have:
$\sqrt{2\sqrt{3-2}} = \sqrt[4]{7 - 3}$
$\sqrt{2\sqrt{1}} = \sqrt[4]{4}$
$\sqrt{2} = \sqrt{2}$
This equation is true, so $t = 3$ is a valid solution.
Checking $t = 19$, we have:
$\sqrt{2\sqrt{19-2}} = \sqrt[4]{7 - 19}$
$\sqrt{2\sqrt{17}} = \sqrt[4]{-12}$
The left side of the equation is a real number, but the right side is not a real number. Therefore, $t = 19$ is not a valid solution.
Thus, the only solution is $t = \boxed{3}$.The answer is: 3
Below is a math problem. Provide a detailed, step-by-step solution.

### Problem:
What is the value of $t$ if $\sqrt{2\sqrt{t-2}} = \sqrt[4]{7 - t}$?

### Answer:
