We are given a system of two equations: $\log y = 2 \log x + 1$ $2y = 9x - 1$ We need to find the values of $x$ and $y$ that satisfy both equations. We assume the logarithm is base 10.

1. Problem Description

We are given a system of two equations:

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

$2y = 9x - 1$

We need to find the values of $x$ and $y$ that satisfy both equations. We assume the logarithm is base

1

0.

2. Solution Steps

From the first equation, we have

$\log y = 2 \log x + 1 = \log (x^2) + \log 10 = \log (10x^2)$

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

$y = 10x^2$

Substituting this into the second equation, we have

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

$20x^2 = 9x - 1$

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

This is a quadratic equation in $x$. We can factor it as follows:

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

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

This gives us two possible solutions for $x$:

$x = \frac{1}{4}$ or $x = \frac{1}{5}$

Now, we can find the corresponding values of $y$ 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 $y = 10 (\frac{1}{5})^2 = 10 (\frac{1}{25}) = \frac{10}{25} = \frac{2}{5}$

We must check if these solutions satisfy the original equations:

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

$2y = 2(\frac{5}{8}) = \frac{5}{4}$

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

So the second equation is satisfied. Also, $x>0$ and $y>0$ so the logarithms are well defined.

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

$2y = 2(\frac{2}{5}) = \frac{4}{5}$

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

So the second equation is satisfied. Also, $x>0$ and $y>0$ so the logarithms are well defined.

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

3. Final Answer

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