1. **State the problem:** Solve the exponential equation $$2^{2x+1} = 3^{x+1}$$. 2. **Recall the properties of exponents:** For any base $a > 0$, $a^{m+n} = a^m \cdot a^n$ and if $a^A = a^B$, then $A = B$ (only if bases are equal). 3. Since the bases 2 and 3 are different, we take the natural logarithm (ln) on both sides to solve for $x$: $$\ln\left(2^{2x+1}\right) = \ln\left(3^{x+1}\right)$$ 4. Use the logarithm power rule $\ln(a^b) = b \ln(a)$: $$ (2x+1) \ln(2) = (x+1) \ln(3) $$ 5. Distribute the logarithms: $$ 2x \ln(2) + \ln(2) = x \ln(3) + \ln(3) $$ 6. Group terms with $x$ on one side and constants on the other: $$ 2x \ln(2) - x \ln(3) = \ln(3) - \ln(2) $$ 7. Factor out $x$: $$ x (2 \ln(2) - \ln(3)) = \ln(3) - \ln(2) $$ 8. Solve for $x$: $$ x = \frac{\ln(3) - \ln(2)}{2 \ln(2) - \ln(3)} $$ 9. This is the exact solution. You can approximate numerically if needed. **Final answer:** $$ x = \frac{\ln(3) - \ln(2)}{2 \ln(2) - \ln(3)} $$