Powerful numbers

[bonanova]

The number abc has a decimal representation of ax10² + bx10¹ + cx10⁰. But let's change things around a bit, and drop the base ten. We could then write abc = a² + b¹ + c⁰ and ask what values of a b c satisfy the equation. There actually may not be a solution. But if we play with the exponents a bit, we might come up with some numbers that do work. Try these

(The numbers are in order, smallest to largest.)

1. ab = a² + b³
2. cd = c² + d³
3. efg = e¹ + f² + g³
4. hij = h¹ + i² + j³
5. klm = k¹ + l² + m³
6. nop = n¹ + o² + p³
7. qrst = q¹ + r² + s³ + t⁴
8. uvwx = u¹ + v² + w³ + x⁴
9. yz@$ = y¹ + z² + @³ + $⁴

There is a shorthand notation. These can also be written ab (2,3), cd (2,3), efg (1,3,5), ... yz@$ (1,2,3,4).

So here's a bonus challenge: abcdefgcc (4, 3, 8, 5, 7, 9, 0, 8, 8)

[Pickett]

Spoiler

1. 43
2. 63
3. 135
4. 175
5. 518
6. 598
7. 1306
8. 1676 <-- Note: v and x have the same value...I'm not sure if that's allowed by the OP.
9. 2427 <-- Note y and @ have the same value...I'm not sure if that's allowed by the OP.

BONUS: Not finding a solution...

      abcdefgcc = a⁴ + b³ + c⁸ + d⁵ + e⁷ + f⁹ + g⁰ + c⁸ + c⁸  OR

      abcdefgcc = a⁴ + b³ + 3c⁸ + d⁵ + e⁷ + f⁹ + 1

Even with allowing duplicate values for the variables and allowing leading zeros in the number on the left, I'm not finding a solution...

 

[bonanova]

Nice work. Bonus solution is the one I had in mind.