We are given two equations: $\log y = 2 \log x + 1$ and $2y = 9x - 1$. We need to solve for $x$ and $y$. We are assuming base 10 logarithm.

1. Problem Description

We are given two equations:

$\log y = 2 \log x + 1$ and $2y = 9x - 1$. We need to solve for $x$ and $y$. We are assuming base 10 logarithm.

2. Solution Steps

First, let's rewrite the first equation using properties of logarithms.

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

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

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

$y = 10x^2$

Now substitute this expression for $y$ into the second equation:

$2y = 9x - 1$

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

$20x^2 = 9x - 1$

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

Now we have a quadratic equation in $x$. We can solve this using the quadratic formula or by factoring. Let's try factoring:

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

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

So the possible values for $x$ are $x = \frac{1}{4}$ or $x = \frac{1}{5}$.

Now we can find the corresponding values for $y$ using $y = 10x^2$:

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

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

Now we check these solutions in the second original equation, $2y = 9x - 1$.

If $x = \frac{1}{4}$ and $y = \frac{5}{8}$, then $2y = 2(\frac{5}{8}) = \frac{5}{4}$ and $9x - 1 = 9(\frac{1}{4}) - 1 = \frac{9}{4} - \frac{4}{4} = \frac{5}{4}$. So this solution works.

If $x = \frac{1}{5}$ and $y = \frac{2}{5}$, then $2y = 2(\frac{2}{5}) = \frac{4}{5}$ and $9x - 1 = 9(\frac{1}{5}) - 1 = \frac{9}{5} - \frac{5}{5} = \frac{4}{5}$. So this solution also works.

3. Final Answer

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