1. **State the problem:** Solve the equation $$3 \log_x 4 + 2 \log_4 x 4 + 3 \log_{16} x \cdot 4 = 0$$ for real numbers $x$. 2. **Rewrite the logarithms:** Recall that $\log_a b = \frac{\log b}{\log a}$ for any positive base $a \neq 1$. 3. **Express each term using a common logarithm (e.g., natural log $\ln$):** $$3 \log_x 4 = 3 \frac{\ln 4}{\ln x}$$ $$2 \log_4 x 4 = 2 \frac{\ln 4}{\ln 4} = 2 \cdot 1 = 2$$ $$3 \log_{16} x \cdot 4 = 3 \frac{\ln x}{\ln 16} \cdot 4 = \frac{12 \ln x}{\ln 16}$$ 4. **Substitute back into the equation:** $$3 \frac{\ln 4}{\ln x} + 2 + \frac{12 \ln x}{\ln 16} = 0$$ 5. **Let $t = \ln x$ (note $x > 0$ and $x \neq 1$ so $t \neq 0$):** $$3 \frac{\ln 4}{t} + 2 + \frac{12 t}{\ln 16} = 0$$ 6. **Multiply both sides by $t$ to clear the denominator:** $$3 \ln 4 + 2t + \frac{12 t^2}{\ln 16} = 0$$ Intermediate step showing cancellation: $$\cancel{t} \left(3 \frac{\ln 4}{\cancel{t}} + 2 + \frac{12 t}{\ln 16}\right) = 0$$ 7. **Multiply both sides by $\ln 16$ to clear the denominator:** $$3 \ln 4 \cdot \ln 16 + 2 t \ln 16 + 12 t^2 = 0$$ 8. **Rewrite as a quadratic in $t$:** $$12 t^2 + 2 t \ln 16 + 3 \ln 4 \cdot \ln 16 = 0$$ 9. **Calculate constants:** $$\ln 4 = \ln(2^2) = 2 \ln 2$$ $$\ln 16 = \ln(2^4) = 4 \ln 2$$ Substitute: $$12 t^2 + 2 t (4 \ln 2) + 3 (2 \ln 2)(4 \ln 2) = 0$$ Simplify: $$12 t^2 + 8 t \ln 2 + 24 (\ln 2)^2 = 0$$ 10. **Divide entire equation by 4:** $$3 t^2 + 2 t \ln 2 + 6 (\ln 2)^2 = 0$$ 11. **Use quadratic formula:** $$t = \frac{-2 \ln 2 \pm \sqrt{(2 \ln 2)^2 - 4 \cdot 3 \cdot 6 (\ln 2)^2}}{2 \cdot 3}$$ Calculate discriminant: $$(2 \ln 2)^2 - 4 \cdot 3 \cdot 6 (\ln 2)^2 = 4 (\ln 2)^2 - 72 (\ln 2)^2 = -68 (\ln 2)^2 < 0$$ 12. **Since the discriminant is negative, there are no real solutions for $t$.** 13. **Recall $t = \ln x$, so no real $x$ satisfies the equation.** **Final answer:** $$\boxed{\text{No real solutions}}$$