To solve the given simultaneous equations, we will work through each equation step by step, using logarithmic properties and algebraic manipulation.

Equation 1:

3 + 4\log_4x = \log_23y

First, we convert the logarithms to the same base. Let's convert everything to base 2 using the change of base formula :

\log_b a = \frac{\log_k a}{\log_k b}

Convert \log_4 x to base 2:

\log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{\log_2 x}{2}

Substitute this into the equation:

3 + 4 \cdot \frac{\log_2 x}{2} = \log_2 3y

Simplify:

3 + 2\log_2 x = \log_2 3y

Express 3 as a logarithm:

\log_2 8 + 2\log_2 x = \log_2 3y

Combine the logarithms on the left:

\log_2 (8x^2) = \log_2 3y

Since the logarithms are equal, the arguments must be equal:

8x^2 = 3y

Equation 2:

\log_3(25x - 3) = 1 + \log_9(y^2)

Convert \log_9(y^2) to base 3:

\log_9(y^2) = \frac{\log_3(y^2)}{\log_3 9} = \frac{2\log_3 y}{2} = \log_3 y

Substitute this into the equation:

\log_3(25x - 3) = 1 + \log_3 y

Express 1 as a logarithm:

\log_3 3 + \log_3 y = \log_3(25x - 3)

Combine the logarithms on the right:

\log_3 (3y) = \log_3(25x - 3)

Since the logarithms are equal, the arguments must be equal:

25x - 3 = 3y

Solving the System of Equations

We now have the system:

8x^2 = 3y
25x - 3 = 3y

Both equations are equal to 3y, so:

25x - 3 = 8x^2

Rearrange to form a quadratic equation:

8x^2 - 25x + 3 = 0

To solve this quadratic equation by factoring, we look for two numbers that multiply to 8 \times 3 = 24 and add to -25. These numbers are -24 and -1.

Rewrite the middle term using these numbers:

8x^2 - 24x - x + 3 = 0

Factor by grouping:

8x(x - 3) - 1(x - 3) = 0

Factor out the common factor (x - 3):

(8x - 1)(x - 3) = 0

Set each factor to zero and solve for x :

8x - 1 = 0 gives x = \frac{1}{8}
x - 3 = 0 gives x = 3

For each x , find y using y=\frac{25x-3}{3}:

If x = 3 :
y=\frac{25\left(3\right)-3}{3}=\frac{72}{3}=24
If x = \frac{1}{8} :
y=\frac{25\left(\frac{1}{8}\right)-3}{3}=\frac{\frac{25}{8}-\frac{24}{8}}{3}=\frac{\frac18}{3}=\frac{1}{24}

Thus, the solutions are:

\boxed{(x,y)=(3,24)\quad\text{and}\quad(x,y)=\left(\frac18,\frac{1}{24}\right)}

Exponents & Logarithms