JAMB 2016 · UTME · Q26✱

Given that $\log_4(y + 1) + \log_4\left(\frac12x\right) = 1$ and $\log_2(y - 1) + \log_2 x = 2$ , solve for $x$ and $y$ respectively.

Worked solution (try it first)

First equation: combine the logs, $\log_4\left(\frac{x(y + 1)}{2}\right) = 1$ , so $\frac{x(y + 1)}{2} = 4$ and $x(y + 1) = 8$ .
Second equation: $\log_2 [x(y - 1)] = 2$ , so $x(y - 1) = 2^2 = 4$ .
Subtract the second from the first: $xy + x - (xy - x) = 8 - 4$ , so $2x = 4$ and $x = 2$ .
Then $2(y + 1) = 8$ gives $y = 3$ .
So $x = 2$ and $y = 3$ , option A.