log_((1)/(2))(x+8)-log_((1)/(2))(x-3)gt log_((1)/(2))(3x)

To solve the inequality $log_{\frac {1}{2}}(x+8)-log_{\frac {1}{2}}(x-3)\gt log_{\frac {1}{2}}(3x)$, we can use the properties of logarithms.<br /><br />First, let's simplify the left side of the inequality using the quotient rule of logarithms:<br /><br />$log_{\frac {1}{2}}(x+8)-log_{\frac {1}{2}}(x-3) = log_{\frac {1}{2}}\left(\frac{x+8}{x-3}\right)$<br /><br />So the inequality becomes:<br /><br />$log_{\frac {1}{2}}\left(\frac{x+8}{x-3}\right) \gt log_{\frac {1}{2}}(3x)$<br /><br />Since the base of the logarithms is $\frac{1}{2}$, which is less than 1, the inequality sign will be reversed when taking the logarithm:<br /><br />$\frac{x+8}{x-3} \lt 3x$<br /><br />Now, let's solve this inequality:<br /><br />$x+8 \lt 3x(x-3)$<br /><br />$x+8 \lt 3x^2 - 9x$<br /><br />$3x^2 - 10x + 8 \lt 0$<br /><br />We can solve this quadratic inequality by factoring:<br /><br />$(3x-4)(x-2) \lt 0$<br /><br />The critical points are $x = \frac{4}{3}$ and $x = 2$. We can test the intervals to find the solution set:<br /><br />For $x \lt \frac{4}{3}$, the inequality is true.<br />For $\frac{4}{3} \lt x \lt 2$, the inequality is false.<br />For $x \gt 2$, the inequality is true.<br /><br />Therefore, the solution set for the inequality $log_{\frac {1}{2}}(x+8)-log_{\frac {1}{2}}(x-3)\gt log_{\frac {1}{2}}(3x)$ is $(-\infty, \frac{4}{3}) \cup (2, \infty)$.