Which is the base? Which is the power?

Here are a few variations that would yield infinitely many solutions:

$A = 16$ , $B = 1$ , and $C = n$ for any integer $n$

$A = 16$ , $B = n$ , and $C = 0$ for any non-zero integer $n$

$A = 16^{n^m}$ , $B = n$ , and $C = -m$ for any non-zero integers $m$ and $n$

For any integer $N,$ $16^{1^N}=2^{2^2},$ which yields the $\boxed{\text{infinitely many}}$ solutions given by $(A,B,C)=(16,1,N)$

$2^{2^2} = 16 = 16^{B^0}$ , where $B$ can be any integer except 0, so long as $A=16$ and $C=0$ . Therefore, there are infinitely many other solutions.

For $(A,B,C) = (2^{20x}, 5x, -1)$

$\large A^{B^C} = {(2^{20x})}^{(5x)^{-1}} = (2^{20x})^{\frac 1{5x}} = 2^{20x \times \frac 1{5x}} = 2^4 = 2^{2^2}$

Where $x$ can be any non-zero integer. Hence there are infinitely many solutions for $(A,B,C)$ .

2^2^2 = 16 = 16^1^n = 16^k^0 where n is any integer and k is any integer NOT equal to 1. So we have infinitely many solutions

For any non-zero integer $n$ ,

$2^{2^2} = 16 = 16^{n^0}$

Therefore there are infinitely many solutions.

For Non-Zero integer Z:

A= 16^Z

B= Z

C= -1

so it is Zth root of 16^Z which is 16

$16^{B^0} = 2^{2^2}$ for all nonzero integers $B$ .

Ultimately we have to get 16. Let's say C=1, so we can forget about it. Now, we know that we can get to any number by putting some number to some power, so we already know, that the answer is infinitely many. For eg. we can take our calculator and input sth like: log840(16) We get: 840^0,4117664097^1=16 So: A=840 B=0,4117664097 C=1

Bcuz any number can be n