Answer
The operation is in base 7.
Explanation
We are asked to determine in what base
[tex]12_b+26_b=41_b[/tex]To do this, we just convert all of these to base 10
[tex]\begin{gathered} 12_b=(1\times b^1)+(2\times b^0)=(1\times b)+(2\times1)=(b+2) \\ 26_b=(2\times b^1)+(6\times b^0)=(2\times b)+(6\times1)=(2b+6) \\ 41_b=(4\times b^1)+(1\times b^0)=(4\times b)+(1\times1)=(4b+1) \end{gathered}[/tex]The equation can then be written in base 10 as
(b + 2) + (2b + 6) = (4b + 1)
b + 2 + 2b + 6 = 4b + 1
b + 2b + 2 + 6 = 4b + 1
3b + 8 = 4b + 1
We can rewrite this as
4b + 1 = 3b + 8
4b - 3b = 8 - 1
b = 7
Hope this Helps!!!
