Solve simultaneously the equations;

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

Answer 1

To solve the given equations simultaneously, we will use the properties of logarithms.

1. Starting with the first equation:
log₂x - log₂y = -1

We can simplify this equation using the logarithmic rule that states that subtracting logarithms with the same base is equivalent to dividing the arguments:
log₂(x/y) = -1

We can rewrite this equation in exponential form using the definition of logarithms:
x/y = 2^(-1)

Simplifying the exponent, we have:
x/y = 1/2

Multiplying both sides by y:
x = y/2

2. Moving on to the second equation:
logy(6x + y) - logy(4x - y) = 1

As in the first equation, we can apply the logarithmic rule of subtracting logarithms to obtain the quotient of the arguments:
logy((6x + y)/(4x - y)) = 1

Converting this equation to exponential form, we have:
(6x + y)/(4x - y) = y^1

Simplifying, we get:
6x + y = 4x - y

Rearranging the equation:
6x - 4x = -y - y
2x = -2y

Dividing both sides by 2:
x = -y

Now, we have obtained two equations:
1. x = y/2
2. x = -y

Combining these equations, we can find the values of x and y:
Since x = y/2 and x = -y, we can equate the two expressions:
y/2 = -y

Multiplying both sides by 2 to eliminate the denominator:
y = -2y

Adding 2y to both sides:
3y = 0

Dividing both sides by 3:
y = 0

Substituting the value of y into either of the initial equations, we can find the value of x:
x = -y = -0 = 0

So, the solution to the given system of equations is x = 0 and y = 0.

Therefore, the correct answer is None of the given options (E) as none of the options include x = 0 and y = 0.

Answer 2

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 $\boxed{\textbf{(A) } x = 0, 2 \text{ and } y = 0, 4}$.