1. **Problem:** Solve the exponential equation $2^{x+1} = 16$. Step 1: Express 16 as a power of 2: $16 = 2^4$. Step 2: Set the exponents equal since the bases are the same: $x+1 = 4$. Step 3: Solve for $x$: $x = 4 - 1 = 3$. 2. **Problem:** Solve the logarithmic equation $\log_3 (x) = 4$. Step 1: Rewrite the logarithmic equation in exponential form: $x = 3^4$. Step 2: Calculate $3^4 = 81$. Step 3: So, $x = 81$. 3. **Problem:** Solve for $x$ in $e^{2x} = 7$. Step 1: Take the natural logarithm of both sides: $\ln(e^{2x}) = \ln(7)$. Step 2: Simplify left side: $2x = \ln(7)$. Step 3: Solve for $x$: $x = \frac{\ln(7)}{2}$. 4. **Problem:** Solve $\log_2 (x-1) + \log_2 (x+3) = 3$. Step 1: Use logarithm property: $\log_2((x-1)(x+3)) = 3$. Step 2: Rewrite in exponential form: $(x-1)(x+3) = 2^3 = 8$. Step 3: Expand: $x^2 + 3x - x - 3 = 8 \Rightarrow x^2 + 2x - 3 = 8$. Step 4: Bring all terms to one side: $x^2 + 2x - 11 = 0$. Step 5: Solve quadratic using the quadratic formula: $x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-11)}}{2} = \frac{-2 \pm \sqrt{4 + 44}}{2} = \frac{-2 \pm \sqrt{48}}{2} = \frac{-2 \pm 4\sqrt{3}}{2} = -1 \pm 2\sqrt{3}$. Step 6: Check domain restrictions: $x-1 > 0 \Rightarrow x > 1$, $x+3 > 0 \Rightarrow x > -3$. Step 7: Only $x = -1 + 2\sqrt{3} \approx 2.464$ satisfies $x > 1$. 5. **Problem:** Solve $5^{2x-1} = \frac{1}{25}$. Step 1: Express $\frac{1}{25}$ as a power of 5: $\frac{1}{25} = 5^{-2}$. Step 2: Set exponents equal: $2x - 1 = -2$. Step 3: Solve for $x$: $2x = -1 \Rightarrow x = -\frac{1}{2}$.