Final answer:
The solved expression is 3^41 / 2^12 after applying the rules of exponents correctly.
Step-by-step explanation:
To solve the expression (3^8 \* 2^-5 \* 9^0)^2 \* (2^-2 / 3^3) \* 3^28, we need to apply the rules of exponents step by step. Remember that when raising an expression inside parentheses to a power, every factor inside is affected by that power. Let's break it down:
- 9^0 = 1 (any number to the power of 0 is 1)
- For the part inside the parentheses: (3^8 \* 2^-5)^2, we first execute the operations within the parentheses and then raise the result to the 2nd power.
- Therefore, inside the parentheses becomes (3^16 \* 2^-10)
- The next segment is (2^-2 / 3^3), division of powers with the same base results in the subtraction of exponents, giving us 2^-2 \* 3^-3.
- Combining all parts: (3^16 \* 2^-10) \* (2^-2 \* 3^-3) \* 3^28
- Add and subtract the exponents for basis 3 and 2 respectively: 3^(16-3+28) \* 2^(-10-2)
- The final result is 3^41 \* 2^-12.
Since we want the answer in positive exponents where possible, we leave 3^41 as it is and write 2^-12 as 1/2^12.
The fully simplified expression is thus 3^41 / 2^12.