Solve the following system of equations: $\log y = 2 \log x + 1$ $2y = 9x - 1$

1. Problem Description

Solve the following system of equations:

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

$2y = 9x - 1$

2. Solution Steps

First equation:

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

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

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

Therefore,

$y = 10x^2$

Second equation:

$2y = 9x - 1$

Substitute the expression for $y$ from the first equation into the second equation:

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

$20x^2 = 9x - 1$

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

Solve the quadratic equation for $x$:

We can factorize the quadratic as follows:

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

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

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

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

If $x = \frac{1}{5}$, then

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

If $x = \frac{1}{4}$, then

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

Check solutions in original equations:

Case 1: $x = \frac{1}{5}$, $y = \frac{2}{5}$

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

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

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

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

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

$\log 2 = \log (\frac{10}{5})$

$\log 2 = \log 2$

The first equation holds.

$2y = 9x - 1$

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

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

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

The second equation holds.

Case 2: $x = \frac{1}{4}$, $y = \frac{5}{8}$

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

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

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

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

$\log 5 - \log 8 = -2 \log 4 + \log 10$

$\log 5 - \log 8 = \log 10 - \log 4^2$

$\log 5 - \log 8 = \log 10 - \log 16$

$\log (\frac{5}{8}) = \log (\frac{10}{16}) = \log (\frac{5}{8})$

The first equation holds.

$2y = 9x - 1$

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

$\frac{10}{8} = \frac{9}{4} - 1$

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

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

The second equation holds.

3. Final Answer

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