We are given two equations: 1) $log(y) = 2log(x) + 1$ 2) $2y = 9x - 1$ We need to solve for $x$ and $y$. We assume that the logarithm is base 10.

1. Problem Description

We are given two equations:

1) $log(y) = 2log(x) + 1$

2) $2y = 9x - 1$

We need to solve for $x$ and $y$. We assume that the logarithm is base

1

0.

2. Solution Steps

First, simplify the logarithmic equation:

$log(y) = 2log(x) + 1$

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

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

Since the logarithms are equal, we can equate the arguments:

$y = 10x^2$

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

$2y = 9x - 1$

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

$20x^2 = 9x - 1$

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

This is a quadratic equation in $x$. We can solve for $x$ using the quadratic formula or by factoring. Let's try factoring:

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

So, $4x - 1 = 0$ or $5x - 1 = 0$

If $4x - 1 = 0$, then $4x = 1$, so $x = \frac{1}{4}$.

If $5x - 1 = 0$, then $5x = 1$, so $x = \frac{1}{5}$.

Now, we find the corresponding values of $y$:

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

Let's check these solutions in the second equation $2y = 9x - 1$:

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

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

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

$\frac{5}{4} = \frac{5}{4}$ (This solution is valid)

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

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

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

$\frac{4}{5} = \frac{4}{5}$ (This solution is valid)

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

3. Final Answer

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