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JAMB 1995 · UME · Q7

Find the value of 1632+log⁡100.0001+log⁡23216^{\frac32} + \log_{10} 0.0001 + \log_2 32.

Worked solution (try it first)
  1. 163216^{\frac32} is the square root of 16, cubed: 43=644^3 = 64.
  2. 0.0001=10−40.0001 = 10^{-4}, so log⁡100.0001=−4\log_{10} 0.0001 = -4.
  3. And 32=2532 = 2^5, so log⁡232=5\log_2 32 = 5.
  4. So the value is 64−4+5=6564 - 4 + 5 = 65, option D.

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