Question: Showing all necessary working, solve the equation $\frac{2e^x+e^{-x}}{e^x+1} = 4$ Giving your answer correct to 3 decimal places.
1. **State the problem:** Solve the equation $$\frac{2e^x+e^{-x}}{e^x+1} = 4$$ for $x$, giving the answer correct to 3 decimal places.
2. **Rewrite the equation:** Let $y = e^x$. Since $e^x > 0$, $y > 0$. The equation becomes:
$$\frac{2y + \frac{1}{y}}{y + 1} = 4$$
3. **Multiply both sides by the denominator:**
$$2y + \frac{1}{y} = 4(y + 1)$$
4. **Multiply both sides by $y$ to clear the fraction:**
$$y(2y + \frac{1}{y}) = y \cdot 4(y + 1)$$
$$2y^2 + 1 = 4y^2 + 4y$$
5. **Bring all terms to one side:**
$$2y^2 + 1 - 4y^2 - 4y = 0$$
$$-2y^2 - 4y + 1 = 0$$
6. **Multiply entire equation by $-1$ to simplify:**
$$2y^2 + 4y - 1 = 0$$
7. **Use the quadratic formula:**
$$y = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{-4 \pm \sqrt{16 + 8}}{4} = \frac{-4 \pm \sqrt{24}}{4}$$
8. **Simplify the square root:**
$$\sqrt{24} = 2\sqrt{6}$$
9. **So:**
$$y = \frac{-4 \pm 2\sqrt{6}}{4} = \frac{-4}{4} \pm \frac{2\sqrt{6}}{4} = -1 \pm \frac{\sqrt{6}}{2}$$
10. **Evaluate the two possible values:**
- $y_1 = -1 + \frac{\sqrt{6}}{2} \approx -1 + 1.2247 = 0.2247$ (valid since $y>0$)
- $y_2 = -1 - \frac{\sqrt{6}}{2} \approx -1 - 1.2247 = -2.2247$ (not valid since $y>0$)
11. **Take natural logarithm to find $x$:**
$$x = \ln(y_1) = \ln(0.2247) \approx -1.492$$
**Final answer:**
$$x \approx -1.492$$ (correct to 3 decimal places)