1. **State the problem:** Solve the equation $3x2^{-x} = 2x3^x$ for $x$. 2. **Rewrite the equation:** The equation is $3x2^{-x} = 2x3^x$. 3. **Divide both sides by $x$ (assuming $x \neq 0$):** $$3 \cdot 2^{-x} = 2 \cdot 3^x$$ 4. **Rewrite the negative exponent:** $$3 \cdot \frac{1}{2^x} = 2 \cdot 3^x$$ 5. **Multiply both sides by $2^x$ to clear the denominator:** $$3 = 2 \cdot 3^x \cdot 2^x = 2 \cdot (3 \cdot 2)^x = 2 \cdot 6^x$$ 6. **Isolate $6^x$:** $$6^x = \frac{3}{2}$$ 7. **Take the natural logarithm of both sides:** $$x \ln(6) = \ln\left(\frac{3}{2}\right)$$ 8. **Solve for $x$:** $$x = \frac{\ln\left(\frac{3}{2}\right)}{\ln(6)}$$ 9. **Check the case $x=0$:** Substitute $x=0$ into original equation: $$3 \cdot 0 \cdot 2^0 = 0$$ $$2 \cdot 0 \cdot 3^0 = 0$$ Both sides equal zero, so $x=0$ is also a solution. **Final answer:** $$x = 0 \quad \text{or} \quad x = \frac{\ln\left(\frac{3}{2}\right)}{\ln(6)}$$