WAEC 2019 · Paper 2 · Q3

(a)
Without using mathematical tables or a calculator, simplify $\dfrac{\log_2 8 + \log_2 16 - 4\log_2 2}{\log_4 16}$ .
(b)
If $1342_{\text{five}} - 241_{\text{five}} = x_{\text{ten}}$ , find the value of $x$ .

Worked solution (try it first)

(a)

Each log is a small whole number: $\log_2 8 = 3$ , $\log_2 16 = 4$ , $\log_2 2 = 1$ and $\log_4 16 = 2$ .
So the expression is $\frac{3 + 4 - 4 \times 1}{2} = \frac32$
$= 1\frac12$ .

(b)

Change both to base ten: $1342_{\text{five}} = 1 \times 125 + 3 \times 25 + 4 \times 5 + 2$
$= 222$ and $241_{\text{five}} = 2 \times 25 + 4 \times 5 + 1$
$= 71$ .
So $x = 222 - 71 = 151$ .

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