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?
 A: First remember that $4^x$ means you multiply $4$ together with itself $x$ times. For example $4^2 = 4*4= 16.$ Note that $4^{400} + 4^{400}+4^{400}+4^{400} = 4^{400}(1+1+1+1) = 4^{400}*4$. And $4^{400}*4$ is just $4$ multiplied together with itself 401 times, which is the same as writing $4^{401}$.
A: $$4^{401}=4^{400+1}=4^{400}\cdot 4^1=4\cdot 4^{400}$$
A: $4^{401} = 4\times 4^{400} = 4^{400} + 4^{400} + 4^{400} + 4^{400}$
A: It only works because $x=401$.
$4^{401}$ means multiply $4$ by itself 401 times.  So $$\begin{array}{rl}
4^{401} &= 4\times 4^{400} \\ &= 4^{400} +4^{400} +4^{400} +4^{400} \end{array}$$
A: $a=4^{400}$
$4^x=a+a+a+a$
$$4^x=4\cdot a$$
$$4^x=4\cdot 4^{400}$$
$$u^m\times u^n= u^{m+n}\implies 4^1\times 4^{400}=4^{401}$$
$$4^x=4^{400+1}$$
$$a^m=a^n \implies m=n$$
$$x=401$$
A: 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
$(4^{400})+(4^{400})+(4^{400})+(4^{400}) = (4^{400})\times 4 = 4^{401}$
