How can $4^x = 4^{400}+4^{400}+4^{400}+4^{400}$ have the solution $x=401$? How can $4^x = 4^{400} + 4^{400} + 4^{400} + 4^{400}$ have the solution $x = 401$?
Can someone explain to me how this works in a simple way?

The number you're adding is being added n (number) times. Well, we can infer from this that if that number is being added n (number) times, it is multiplicating itself. Now, if a number is multiplicating itself, then we have an exponentiation!
$N_1+N_2+N_3+...+N_N = N\times N = N^2$
If you're adding a number, no matter how big it is or what operation you're doing with it, and you're adding it n (number) times, you'll end up with
$(N^k)1+(N^k)2+(N^k)3+...(N^k)N = (N^k)N$
Which is  $N^{k+1}$
So, if you're doing it with 4^400, we've got
$(4^{400})+(4^{400})+(4^{400})+(4^{400}) = (4^{400})\times 4 = 4^{401}$
