We are given two equations: 1. $\log y = 2 \log x + 1$

1. Problem Description

We are given two equations:

1. $\log y = 2 \log x + 1$

2. $2y = 9x - 1$

We need to solve for $x$ and $y$. Here the base of the logarithm is assumed to be

1

0.

2. Solution Steps

Step 1: Simplify the logarithmic equation.

Using the property of logarithms, $a \log b = \log b^a$, we can rewrite the first equation as:

$\log y = \log x^2 + 1$

Since $1 = \log 10$, we have

$\log y = \log x^2 + \log 10$

Using the property $\log a + \log b = \log (ab)$, we get:

$\log y = \log (10x^2)$

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

$y = 10x^2$

Step 2: Substitute the expression for $y$ into the second equation.

We have $2y = 9x - 1$. Substituting $y = 10x^2$ into this equation, we get:

$2(10x^2) = 9x - 1$

$20x^2 = 9x - 1$

$20x^2 - 9x + 1 = 0$

Step 3: Solve the quadratic equation.

We can solve the quadratic equation $20x^2 - 9x + 1 = 0$ by factoring:

$20x^2 - 5x - 4x + 1 = 0$

$5x(4x - 1) - 1(4x - 1) = 0$

$(5x - 1)(4x - 1) = 0$

This gives us two possible solutions for $x$:

$5x - 1 = 0 \implies x = \frac{1}{5}$

$4x - 1 = 0 \implies x = \frac{1}{4}$

Step 4: Find the corresponding values of $y$.

For $x = \frac{1}{5}$:

$y = 10x^2 = 10(\frac{1}{5})^2 = 10(\frac{1}{25}) = \frac{10}{25} = \frac{2}{5}$

For $x = \frac{1}{4}$:

$y = 10x^2 = 10(\frac{1}{4})^2 = 10(\frac{1}{16}) = \frac{10}{16} = \frac{5}{8}$

Step 5: Check the solutions with the original equations.

For $x = \frac{1}{5}$ and $y = \frac{2}{5}$:

Equation 1: $\log y = 2 \log x + 1$

$\log (\frac{2}{5}) = 2 \log (\frac{1}{5}) + 1$

$\log (\frac{2}{5}) = 2(-\log 5) + 1$

$\log 2 - \log 5 = -2 \log 5 + \log 10$

$\log 2 - \log 5 = -2 \log 5 + \log 2 + \log 5$

$\log 2 - \log 5 = \log 2 - \log 5$, which is true.

Equation 2: $2y = 9x - 1$

$2(\frac{2}{5}) = 9(\frac{1}{5}) - 1$

$\frac{4}{5} = \frac{9}{5} - \frac{5}{5}$

$\frac{4}{5} = \frac{4}{5}$, which is true.

For $x = \frac{1}{4}$ and $y = \frac{5}{8}$:

Equation 1: $\log y = 2 \log x + 1$

$\log (\frac{5}{8}) = 2 \log (\frac{1}{4}) + 1$

$\log 5 - \log 8 = 2(-\log 4) + 1$

$\log 5 - 3 \log 2 = -2(2 \log 2) + \log 10$

$\log 5 - 3 \log 2 = -4 \log 2 + \log 2 + \log 5$

$\log 5 - 3 \log 2 = -3 \log 2 + \log 5$, which is true.

Equation 2: $2y = 9x - 1$

$2(\frac{5}{8}) = 9(\frac{1}{4}) - 1$

$\frac{5}{4} = \frac{9}{4} - \frac{4}{4}$

$\frac{5}{4} = \frac{5}{4}$, which is true.

3. Final Answer

The solutions are:

$(x, y) = (\frac{1}{5}, \frac{2}{5})$ and $(x, y) = (\frac{1}{4}, \frac{5}{8})$.