Question
Solve simultaneously the equations;
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
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
Answers
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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 $\boxed{\textbf{(A) } x = 0, 2 \text{ and } y = 0, 4}$.
$$\log_2 \frac{x}{y} = -1$$
$$\frac{x}{y} = \frac{1}{2}$$
$$x = \frac{y}{2}$$
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 $\boxed{\textbf{(A) } x = 0, 2 \text{ and } y = 0, 4}$.