WAEC 2019 · Paper 2 · Q6

(a)
Given that $2^m \times \left(\frac18\right)^n = 128$ and $4^m \div 2^{-4n} = \frac{1}{16}$ , find the value of $(m - n)$ .
(b)
Find the equation of the line joining the points $\left(-2, \frac12\right)$ and $\left(1, -\frac23\right)$ .
Show the answer
$18y + 7x + 5 = 0$

Worked solution (try it first)

(a)

In base 2: $2^m \times (2^{-3})^n = 2^7$ , so $m - 3n = 7$ .
And $4^m \div 2^{-4n} = 2^{2m} \times 2^{4n}$
$= 2^{-4}$ , so $2m + 4n = -4$ , i.e. $m + 2n = -2$ .
Subtract the second equation from the first: $-5n = 9$ , so $n = -\frac95$ .
Then $m = 7 + 3n = 7 - \frac{27}{5} = \frac85$ .
$m - n = \frac85 + \frac95$
$= \frac{17}{5}$
$= 3\frac25$ .

(b)

Gradient $= \frac{-\frac23 - \frac12}{1 - (-2)}$
$= \frac{-\frac76}{3}$
$= -\frac{7}{18}$ .
Through $\left(-2, \frac12\right)$ : $y - \frac12 = -\frac{7}{18}(x + 2)$ .
Multiply by 18: $18y - 9 = -7x - 14$ , so $18y + 7x + 5 = 0$ .

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