1. **State the problem:** Show that $\left(\log_2 3\right)\left(\log_3 4\right) = 2$. 2. **Recall the change of base formula:** For any positive numbers $a,b,c$ with $a \neq 1$ and $b \neq 1$, $$\log_a b = \frac{\log_c b}{\log_c a}.$$ This allows us to express logarithms in terms of a common base. 3. **Express each logarithm in terms of base 2:** $$\log_2 3 = \log_2 3$$ $$\log_3 4 = \frac{\log_2 4}{\log_2 3}$$ 4. **Substitute into the product:** $$\left(\log_2 3\right)\left(\log_3 4\right) = \left(\log_2 3\right) \times \frac{\log_2 4}{\log_2 3}$$ 5. **Cancel common factors:** $$= \cancel{\log_2 3} \times \frac{\log_2 4}{\cancel{\log_2 3}} = \log_2 4$$ 6. **Evaluate $\log_2 4$:** Since $4 = 2^2$, $$\log_2 4 = 2$$ 7. **Conclusion:** $$\left(\log_2 3\right)\left(\log_3 4\right) = 2$$ This completes the proof.