NECO 2023 · Paper 1 · Q4

Indices, logarithms & surds

Simplify $\frac12\log_4 64 + \log_4 16 - \log_4 2$ .

Worked solution (try it first)

$64 = 4^3$ , so $\log_4 64 = 3$ and $\frac12\log_4 64 = \frac32$ .
$16 = 4^2$ , so $\log_4 16 = 2$ .
And $2 = 4^{1/2}$ , so $\log_4 2 = \frac12$ .
So the value is $\frac32 + 2 - \frac12 = 3$ , option B.

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