WAEC 2019 · Paper 2 · Q9✱✱

(a)
Solve the equations $3\log_2 x = y$ and $\log_2 4x = y + 4$ simultaneously.
Show the answer
$x = \frac12$ , $y = -3$
(b)
A binary operation $*$ is defined on $\mathbb R$ by $a * b = a^2 - 2ab + b^2$ . If $(-3) * 5 = 2^n$ , find $n$ .

Worked solution (try it first)

Change each equation to index form: $3\log_2 x = y$ gives $2^y = x^3$ , and $\log_2 4x = y + 4$ gives $2^{y + 4} = 4x$ .
$2^{y + 4} = 16 \times 2^y = 16x^3$ , so $16x^3 = 4x$ .
Rearrange: $16x^3 - 4x = 0$ , so $4x(4x^2 - 1) = 0$ and $x = 0$ or $x = \pm\frac12$ .
Only $x = \frac12$ has a logarithm ( $\log_2 0$ and $\log_2(-\frac12)$ don't exist).
Then $y = 3\log_2 \frac12 = 3(-1) = -3$ .
So $x = \frac12$ , $y = -3$ .