2022/056) What is the value of a, b, c If $2^a+4^b+8^c=328$ and a, b, c are natural numbers?

 Let us express 328 as sum of power of 2 (as 2,4,8 all are power of 2)

$328 = 256 + 64 + 8 = 2^8 + 2^ 6 + 2^3$

Only $2^6$ or $2^3$ are power of 8

So let us consider these 2 cases

1) $2^6 = 8^2$ and 8 is not power of 4 so $8 = 2^3$ and $256=4^4$ giving a = 3, b = 4, c= 1

2) $2^3 = 8^1$ and this gives 2 cases 

        case 1 $4^4 = 256, 2^6 = 64$ giving a = 6, b = 4, c = 1

        case 2  $2^8 = 256, 4^3 = 64$ giving a = 8, b = 3, c = 1

The above 3 are solutions