1. The problem is to find the sign of the function $f(x) = \ln\left(\frac{2x-3}{3x+4}\right)$.\n\n2. Recall that the natural logarithm function $\ln(y)$ is defined only for $y > 0$, and its sign depends on whether $y$ is greater than 1 or between 0 and 1:\n- If $y > 1$, then $\ln(y) > 0$.\n- If $0 < y < 1$, then $\ln(y) < 0$.\n\n3. So we need to analyze the sign of the fraction $\frac{2x-3}{3x+4}$ and determine when it is greater than 1, between 0 and 1, or not positive.\n\n4. First, find the domain where the argument is positive:\n$$\frac{2x-3}{3x+4} > 0.$$\nThis inequality holds when numerator and denominator have the same sign.\n\n5. Solve $2x-3 > 0 \Rightarrow x > \frac{3}{2}$.\nSolve $3x+4 > 0 \Rightarrow x > -\frac{4}{3}$.\n\n6. The fraction is positive if:\n- Both numerator and denominator are positive: $x > \frac{3}{2}$, or\n- Both numerator and denominator are negative: $x < -\frac{4}{3}$.\n\n7. Next, find when the fraction is greater than 1:\n$$\frac{2x-3}{3x+4} > 1.$$\nMultiply both sides by $3x+4$ (consider sign):\n\nCase 1: $3x+4 > 0 \Rightarrow x > -\frac{4}{3}$\n$$2x - 3 > 3x + 4 \Rightarrow -x > 7 \Rightarrow x < -7.$$\nBut $x > -\frac{4}{3}$ contradicts $x < -7$, so no solution here.\n\nCase 2: $3x+4 < 0 \Rightarrow x < -\frac{4}{3}$\nMultiply inequality by negative number, flip sign:\n$$2x - 3 < 3x + 4 \Rightarrow -x < 7 \Rightarrow x > -7.$$\nCombined with $x < -\frac{4}{3}$, solution is $-7 < x < -\frac{4}{3}$.\n\n8. Find when the fraction is between 0 and 1:\n$$0 < \frac{2x-3}{3x+4} < 1.$$\nFrom step 6, positive intervals are $x < -\frac{4}{3}$ or $x > \frac{3}{2}$.\nFrom step 7, fraction $>1$ only on $-7 < x < -\frac{4}{3}$.\nSo for $x > \frac{3}{2}$, check if fraction $<1$:\n$$\frac{2x-3}{3x+4} < 1.$$\nMultiply both sides by $3x+4 > 0$:\n$$2x - 3 < 3x + 4 \Rightarrow -x < 7 \Rightarrow x > -7,$$\nwhich is true for $x > \frac{3}{2}$.\nSo for $x > \frac{3}{2}$, fraction is positive and less than 1.\n\n9. Summary of sign of $f(x)$:\n- For $x < -\frac{4}{3}$, fraction positive and $-7 < x < -\frac{4}{3}$ fraction $>1$ so $f(x) > 0$ on $(-7, -\frac{4}{3})$.\n- For $x < -7$, fraction positive but less than 1, so $f(x) < 0$ on $(-\infty, -7)$.\n- For $x > \frac{3}{2}$, fraction positive and less than 1, so $f(x) < 0$ on $(\frac{3}{2}, \infty)$.\n- For $-\frac{4}{3} < x < \frac{3}{2}$, fraction negative or undefined, so $f(x)$ undefined.\n\n10. Final answer: The sign of $f(x)$ is positive on $(-7, -\frac{4}{3})$ and negative on $(-\infty, -7)$ and $(\frac{3}{2}, \infty)$, undefined elsewhere.