Solve the Logarithmic Inequality: log₂3 - log₂(x+3) ≤ 8

$\log_23-\log_2(x+3)\le8$

To solve this problem, we'll apply the properties of logarithms and inequality manipulation.

Initially, consider the given inequality:

$\log_2 3 - \log_2 (x + 3) \le 8$

Using the quotient rule of logarithms, combine the logs:

$\log_2 \left(\frac{3}{x + 3}\right) \le 8$

The inequality $\log_2 \left(\frac{3}{x + 3}\right) \le 8$ can be rewritten by converting the logarithm to an exponential form:

$\frac{3}{x + 3} \le 2^8$

Since $2^8 = 256$, substitute to get:

$\frac{3}{x + 3} \le 256$

To remove the fraction, multiply both sides by $x + 3$, assuming $x + 3 > 0$ to maintain the inequality direction:

$3 \le 256(x + 3)$

Divide by 256 to isolate $x+3$:

$\frac{3}{256} \le x + 3$

Subtract 3 from both sides to solve for $x$:

$x \ge \frac{3}{256} - 3$

Given the problem's constraints about the positivity of the logarithm's argument, ensure $x > -3$. Our derived inequality starts from $x \ge \frac{3}{256} - 3$, which satisfies this, thus correctly addressing the domain assumptions.

In conclusion, the solution to the inequality is:

$x \ge \frac{3}{256} - 3$