Solve simultaneously the equations;

Question

log2 𝑥 − log2 𝑦 = −1 and
log𝑦(6𝑥 + 𝑦) − log𝑦(4𝑥 − 𝑦) = 1
A. 𝑥 = 0, 2 𝑎𝑛𝑑 𝑦 = 0, 4
B. 𝑥 = −2, 0 𝑎𝑛𝑑 𝑦 = −4, 0
C. 𝑥 = 0, 4 𝑎𝑛𝑑 𝑦 = 0, 2
D. 𝑥 = −4, 0 𝑎𝑛𝑑 𝑦 = −2, 0

Answer 1

We can use the logarithmic identity $\log a - \log b = \log \frac{a}{b}$ to simplify the first equation:

$$\log_2 \frac{x}{y} = -1$$

$$\frac{x}{y} = \frac{1}{2}$$

$$x = \frac{y}{2}$$

Using the second equation and the fact that $\log_a (bc) = \log_a b + \log_a c$, we get:

$$\log_y \frac{(6x+y)}{(4x-y)} = 1$$

$$\frac{(6x+y)}{(4x-y)} = y$$

$$6x+y = 4xy-y^2$$

Substituting $x=\frac{y}{2}$:

$$6\left(\frac{y}{2}\right)+y = 4\left(\frac{y}{2}\right)y-y^2$$

$$3y+y=2y^2-y^2$$

$$4y=y^2$$

$$y^2-4y=0$$

$$y(y-4)=0$$

So $y=0$ or $y=4$. If $y=0$, then $x=\frac{y}{2}=0$, but this is not a valid solution since we cannot take the logarithm of 0. If $y=4$, then $x=\frac{y}{2}=2$. Therefore, the solution is $\boxed{\textbf{(A) } x = 0, 2 \text{ and } y = 0, 4}$.