Find the value of x for which \(32_{four} = 22_x\)

  • A three
  • B five
  • C six
  • D seven

The correct answer is C. six

\(32_4 = 22_x\)

\(3 \times 4^1 + 2 \times 4^o\) = \(2 \times x^1 + 2 \times x^o\)

12 + 2 x 1 = 2x + 2 x 1

14 = 2x + 2

14 - 2 = 2x

12 = 2x

x = \(\frac{12}{2}\)

x = 6

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