### Instruction: Solve over the integers: $$2^a+4^b+8^c=328.$$.### Response: I notice that all the terms on the left side of the equation are powers of 2, so I wonder if I can rewrite the equation in terms of a single base.
I recall that $4=2^2$ and $8=2^3$, so I can substitute those expressions into the equation and use the exponent rules to get: $$2^a+2^{2b}+2^{3c}=328.$$
Now I have a single base on the left side, but the right side is not a power of 2.
However, I know that every positive integer can be written as a sum of distinct powers of 2, so I try to find such a representation for 328.
I start with the largest power of 2 that is less than 328, which is 256.
Then I subtract 256 from 328 and get 72.
I repeat the process with 72 and find the next largest power of 2, which is 64.
Subtracting 64 from 72 gives me 8, which is also a power of 2.
So I have found that $$328=256+64+8=2^8+2^6+2^3.$$
I can now equate the two sides of the equation and use the fact that if $2^x=2^y$, then $x=y$ to get a system of three equations: $$\begin{aligned} 2^a&=2^8 \\ 2^{2b}&=2^6 \\ 2^{3c}&=2^3 \end{aligned}$$
Solving for $a$, $b$, and $c$ is easy now, I just need to divide the exponents by the coefficients of the variables.