To convert the numeral [tex]\(35,451_8\)[/tex] (a base-8 number) to its equivalent in base-10, we need to understand the place value system for base-8.
Each digit in the base-8 numeral represents a power of 8. The rightmost digit represents [tex]\(8^0\)[/tex], the next represents [tex]\(8^1\)[/tex], then [tex]\(8^2\)[/tex], and so on. In general, for a numeral [tex]\(d_4d_3d_2d_1d_0\)[/tex] in base-8, the base-10 equivalent can be found by:
[tex]\[ d_4 \times 8^4 + d_3 \times 8^3 + d_2 \times 8^2 + d_1 \times 8^1 + d_0 \times 8^0 \][/tex]
Given the numeral [tex]\(35,451_8\)[/tex]:
- The leftmost digit [tex]\(3\)[/tex] is in the [tex]\(8^4\)[/tex] place.
- The next digit [tex]\(5\)[/tex] is in the [tex]\(8^3\)[/tex] place.
- The digit [tex]\(4\)[/tex] is in the [tex]\(8^2\)[/tex] place.
- The digit [tex]\(5\)[/tex] is in the [tex]\(8^1\)[/tex] place.
- The rightmost digit [tex]\(1\)[/tex] is in the [tex]\(8^0\)[/tex] place.
We substitute and calculate:
[tex]\[ 35,451_8 = 3 \times 8^4 + 5 \times 8^3 + 4 \times 8^2 + 5 \times 8^1 + 1 \times 8^0 \][/tex]
Now, we evaluate each term:
- [tex]\(3 \times 8^4 = 3 \times 4096 = 12288\)[/tex]
- [tex]\(5 \times 8^3 = 5 \times 512 = 2560\)[/tex]
- [tex]\(4 \times 8^2 = 4 \times 64 = 256\)[/tex]
- [tex]\(5 \times 8^1 = 5 \times 8 = 40\)[/tex]
- [tex]\(1 \times 8^0 = 1 \times 1 = 1\)[/tex]
Next, we add these individual results together:
[tex]\[ 12288 + 2560 + 256 + 40 + 1 = 15145 \][/tex]
Thus, the base-10 equivalent of [tex]\(35,451_8\)[/tex] is:
[tex]\[ 35,451_8 = 15145_{10} \][/tex]
