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JAMB 1993 · UME · Q4✱✱

Solve for yy in the equation 10y×52y−2×4y−1=110^y \times 5^{2y - 2} \times 4^{y - 1} = 1.

Worked solution (try it first)
  1. Write each factor with bases 2 and 5: 10y=2y×5y10^y = 2^y \times 5^y and 4y−1=22y−24^{y - 1} = 2^{2y - 2}.
  2. Collect the powers: 2y+2y−2×5y+2y−2=23y−2×53y−22^{y + 2y - 2} \times 5^{y + 2y - 2} = 2^{3y - 2} \times 5^{3y - 2}, which is 103y−210^{3y - 2}.
  3. 103y−2=1=10010^{3y - 2} = 1 = 10^0, so 3y−2=03y - 2 = 0 and y=23y = \frac23, option B.

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