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JAMB 2003 · UME · Q7

Evaluate log⁡24+log⁡1216−log⁡432\log_{\sqrt2} 4 + \log_{\frac12} 16 - \log_4 32.

Worked solution (try it first)
  1. (2)4=4(\sqrt2)^4 = 4, so log⁡24=4\log_{\sqrt2} 4 = 4.
  2. (12)−4=16\left(\frac12\right)^{-4} = 16, so log⁡1216=−4\log_{\frac12} 16 = -4.
  3. 42.5=25=324^{2.5} = 2^5 = 32, so log⁡432=2.5\log_4 32 = 2.5.
  4. So the value is 4+(−4)−2.5=−2.54 + (-4) - 2.5 = -2.5, option A.

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