A number system uses $7$ as its base. In such system, $2^{10} =$?

Question

Basically, the question asks what is:

$(2^{10})_7 = x_7$ Where x is an integer.

Now I have two questions:

1. Is there a property wherein $(A^x)_y = (A^y)_x$? If so, then please explain how that's possible ?
2. Had this question been $(2^8)_7 = x_7$, how would I solve it?

Answer 1

There is a difference between $(2^{10})_7$ and $(2^{10_7})_7$.

The first is subject to different interpretations, depending on whether "$10$" is base 10 or base 7.

If base 7, then $10_7 = 7_{10}$ so the value is $(2^{10_7})_7 =(2^{7_{10}})_7 =(128_{10})_7 =2x49_{10}+4x7+2 =242_7 $.

Answer 2

Assuming all my numbers are in base 10, (as @marty cohen points out we need to specify) $$2^{10} = 2 *(7+1)^3$$ You can then use the binomial theorem to expand the cube in powers of $7$ and then just do a little clean up for powers where the co-efficient exceeds $7$.