1. **State the problem:** Solve the equation $$e^{3x} = 7.5 + x$$ for $x$. 2. **Understand the equation:** This is a transcendental equation because it involves both an exponential function and a linear term in $x$. Such equations usually cannot be solved exactly using elementary algebraic methods. 3. **Approach:** We can try to solve it using numerical methods or graphically. However, to understand the behavior, let's analyze it. 4. **Rewrite the equation:** Define a function $$f(x) = e^{3x} - (7.5 + x)$$. We want to find $x$ such that $$f(x) = 0$$. 5. **Check values to locate root:** - At $x=0$, $$f(0) = e^0 - (7.5 + 0) = 1 - 7.5 = -6.5$$ (negative) - At $x=1$, $$f(1) = e^3 - (7.5 + 1) \approx 20.0855 - 8.5 = 11.5855$$ (positive) Since $f(0)<0$ and $f(1)>0$, by the Intermediate Value Theorem, there is at least one root between $0$ and $1$. 6. **Use Newton's method for approximation:** - The derivative is $$f'(x) = 3e^{3x} - 1$$. - Start with initial guess $x_0 = 0.5$. 7. **Iteration formula:** $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$ 8. **Calculate:** - $f(0.5) = e^{1.5} - (7.5 + 0.5) \approx 4.4817 - 8 = -3.5183$ - $f'(0.5) = 3e^{1.5} - 1 \approx 3 \times 4.4817 - 1 = 13.4451$ $$x_1 = 0.5 - \frac{-3.5183}{13.4451} = 0.5 + 0.2618 = 0.7618$$ 9. **Next iteration:** - $f(0.7618) = e^{2.2854} - (7.5 + 0.7618) \approx 9.829 - 8.2618 = 1.5672$ - $f'(0.7618) = 3e^{2.2854} - 1 \approx 3 \times 9.829 - 1 = 28.487$ $$x_2 = 0.7618 - \frac{1.5672}{28.487} = 0.7618 - 0.055 = 0.7068$$ 10. **Next iteration:** - $f(0.7068) = e^{2.1204} - (7.5 + 0.7068) \approx 8.337 - 8.2068 = 0.1302$ - $f'(0.7068) = 3e^{2.1204} - 1 \approx 3 \times 8.337 - 1 = 24.011$ $$x_3 = 0.7068 - \frac{0.1302}{24.011} = 0.7068 - 0.0054 = 0.7014$$ 11. **Next iteration:** - $f(0.7014) = e^{2.1042} - (7.5 + 0.7014) \approx 8.201 - 8.2014 = -0.0004$ Close to zero, so solution is approximately $$x \approx 0.7014$$. **Final answer:** $$x \approx 0.7014$$