Solve the Logarithmic Inequality: ln(x+5) + ln(x) ≤ ln(4) + ln(2x)

To solve this problem, we'll follow these steps:

• Step 1: Use properties of logarithms to combine terms.

• Step 2: Transform the logarithmic inequality into an algebraic form.

• Step 3: Solve the resulting inequality.

• Step 4: Check the domain restrictions and verify the solution.

Let's work through each step:

Step 1: Use the property $\ln a + \ln b = \ln(ab)$:
$\ln(x+5) + \ln x = \ln((x+5)x) = \ln(x^2 + 5x)$
$\ln 4 + \ln 2x = \ln(4 \cdot 2x) = \ln(8x)$

Step 2: Set up the inequality:
$\ln(x^2 + 5x) \le \ln(8x)$

Step 3: Since the logarithmic functions are equal (i.e., both ordinals are decreasing or increasing simultaneously), we can drop logarithms (as long as the arguments are positive):
$x^2 + 5x \le 8x$
Simplify the inequality to:
$x^2 + 5x - 8x \le 0$
$x^2 - 3x \le 0$

Step 4: Factor the quadratic inequality:
$x(x - 3) \le 0$

Determine the critical points of the expression by setting each factor to zero:
$x = 0 \text{ and } x = 3$

The critical points divide the number line into intervals: x < 0 , 0 \le x < 3 , and x > 3 . Test these intervals:

• For x < 0 , pick $x = -1$; the expression $(-1)(-1 - 3) = -4$, which is not less than or equal to zero.

• For 0 < x < 3 , pick $x = 1$; the expression $1(1 - 3) = -2$, which satisfies the inequality.

• For x > 3 , pick $x = 4$; the expression $4(4 - 3) = 4$, which does not satisfy the inequality.

Finally, consider the endpoints:

• At $x = 0$, the inequality does not hold due to the logarithm constraints (undefined).

• At $x = 3$, substitute $x$ into the simplified inequality: $3(3 - 3) = 0$, which satisfies the inequality.

Therefore, $x$ must satisfy the inequality 0 < x \le 3 to maintain positive arguments for the logarithms and satisfy the inequality.

Thus, the solution to the problem is 0 < x \le 3 , or choice 2.