1. **State the problem:** Solve the equation $9^x - 6^x = 4^x$ for $x$. 2. **Rewrite the bases as powers of primes:** $$9 = 3^2, \quad 6 = 2 \times 3, \quad 4 = 2^2$$ So the equation becomes: $$ (3^2)^x - (2 \times 3)^x = (2^2)^x $$ which simplifies to: $$ 3^{2x} - (2^x \cdot 3^x) = 2^{2x} $$ 3. **Introduce substitutions:** Let $a = 3^x$ and $b = 2^x$. Then the equation is: $$ a^2 - ab = b^2 $$ 4. **Rewrite the equation:** $$ a^2 - ab - b^2 = 0 $$ 5. **Divide both sides by $b^2$ (assuming $b \neq 0$):** $$ \frac{a^2}{b^2} - \frac{ab}{b^2} - \frac{b^2}{b^2} = 0 $$ which is: $$ \left(\frac{a}{b}\right)^2 - \frac{a}{b} - 1 = 0 $$ Using cancellation notation: $$ \cancel{b^2} \left(\frac{a}{\cancel{b}}\right)^2 - \frac{a}{\cancel{b}} - 1 = 0 $$ 6. **Let $y = \frac{a}{b} = \frac{3^x}{2^x} = \left(\frac{3}{2}\right)^x$. The quadratic equation is:** $$ y^2 - y - 1 = 0 $$ 7. **Solve the quadratic equation using the quadratic formula:** $$ y = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2} $$ 8. **Since $y = \left(\frac{3}{2}\right)^x > 0$, we take the positive root:** $$ y = \frac{1 + \sqrt{5}}{2} $$ 9. **Solve for $x$:** $$ \left(\frac{3}{2}\right)^x = \frac{1 + \sqrt{5}}{2} $$ Take the natural logarithm of both sides: $$ x \ln\left(\frac{3}{2}\right) = \ln\left(\frac{1 + \sqrt{5}}{2}\right) $$ 10. **Isolate $x$:** $$ x = \frac{\ln\left(\frac{1 + \sqrt{5}}{2}\right)}{\ln\left(\frac{3}{2}\right)} $$ **Final answer:** $$ x = \frac{\ln\left(\frac{1 + \sqrt{5}}{2}\right)}{\ln\left(\frac{3}{2}\right)} $$