We are given two equations: $log(y) = 2log(x) + 1$ $2y = 9x - 1$ 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 two equations:

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

$2y = 9x - 1$

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

First, we rewrite the first equation using properties of logarithms.

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

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

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

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

$y = 10x^2$

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

$2y = 9x - 1$

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

$20x^2 = 9x - 1$

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

We can factor this quadratic equation:

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

So the possible values for $x$ are:

$4x - 1 = 0 => x = 1/4$

$5x - 1 = 0 => x = 1/5$

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

If $x = 1/4$, then $y = 10(1/4)^2 = 10(1/16) = 10/16 = 5/8$

If $x = 1/5$, then $y = 10(1/5)^2 = 10(1/25) = 10/25 = 2/5$

We now check if these solutions satisfy the second equation $2y = 9x - 1$:

If $x = 1/4$ and $y = 5/8$, then $2y = 2(5/8) = 5/4$ and $9x - 1 = 9(1/4) - 1 = 9/4 - 4/4 = 5/4$. This solution works.

If $x = 1/5$ and $y = 2/5$, then $2y = 2(2/5) = 4/5$ and $9x - 1 = 9(1/5) - 1 = 9/5 - 5/5 = 4/5$. This solution works.

3. Final Answer

The solutions are $(x, y) = (1/4, 5/8)$ and $(x, y) = (1/5, 2/5)$.