Given that \(\log_{3}(x - y) = 1\) and \(\log_{3}(2x + y) = 2\), find the value of x.

Given that \(\log_{3}(x - y) = 1\) and \(\log_{3}(2x + y) = 2\), find the value of x.

Answer Details

We can use the properties of logarithms to simplify the given expressions and then solve for x. Firstly, from the first equation, we have: \[\log_{3}(x - y) = 1\] \[x - y = 3\] Similarly, from the second equation, we have: \[\log_{3}(2x + y) = 2\] \[2x + y = 9\] Now, we have a system of two equations with two variables, x and y. We can solve for y by subtracting the first equation from the second: \[(2x + y) - (x - y) = 9 - 3\] \[x + 2y = 6\] \[y = 3 - \frac{1}{2}x\] Substituting this value of y into the first equation: \[x - (3 - \frac{1}{2}x) = 3\] \[\frac{3}{2}x = 6\] \[x = 4\] Therefore, the value of x is 4, which is.