WAEC 2011 · Paper 2 · Q3

(a)
Solve for $x$ and $y$ in the equations: $\log(x - 1) + 2\log y = 2\log3$ ; $\log x + \log y = \log6$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

First equation: $2\log y = \log y^2$ , so the left side is $\log[(x - 1)y^2]$ .
And $2\log 3 = \log 9$ .
So $(x - 1)y^2 = 9$ .
Second equation: $\log x + \log y = \log(xy)$ , so $xy = 6$ , which gives $x = \frac6y$ .
Substitute into the first: $\left(\frac6y - 1\right)y^2 = 9$ .
Multiply out: $6y - y^2 = 9$ , so $y^2 - 6y + 9 = 0$ .
Factorise: $(y - 3)^2 = 0$ , so $y = 3$ .
Then $x = \frac63 = 2$ .
Check: $\log 1 + 2\log 3 = 2\log 3$ ✓ and $\log 2 + \log 3 = \log 6$ ✓.
So $x = 2$ , $y = 3$ .