Find the value of x for which \(32_{four} = 22_x\)
- three
- five
-
six
- seven
The correct answer is: C
Explanation
\(32_4 = 22_x\)
\(3 \times 4^1 + 2 \times 4^o\) = \(2 \times x^1 + 2 \times x^o\)
12 + 2 x 1 = 2x + 2 x 1
14 = 2x + 2
14 - 2 = 2x
12 = 2x
x = \(\frac{12}{2}\)
x = 6