To solve this problem using the Bisection Method, we find the root of $f(t) = e^{-x} - 3\log x$ within the interval $[x_a, x_b] = [0.5, 1.5]$, with a tolerance $\varepsilon = 0.01$.

Steps of the Bisection Method:

1. Define the function: $f(t) = e^{-x} - 3\log x$.

2. Check initial values: Evaluate $f(0.5)$ and $f(1.5)$ to confirm the root lies in $[0.5, 1.5]$. The sign of $f(0.5)$ and $f(1.5)$ must differ.

3. Compute the midpoint: $c = \frac{x_a + x_b}{2}$, and evaluate $f(c)$.

4. Update the interval: If $f(c)$ has the same sign as $f(x_a)$, set $x_a = c$; otherwise, set $x_b = c$.

5. Check for convergence: Stop if $|x_b - x_a| < \varepsilon$, or $|f(c)| < \varepsilon$.

Let’s calculate step by step.

Calculation:

1. Initial evaluations: $f(0.5) = e^{-0.5} - 3\log(0.5) = 0.6065 - (-2.0794) = 2.6859 > 0,$ $f(1.5) = e^{-1.5} - 3\log(1.5) = 0.2231 - 1.2164 = -0.9933 < 0.$ Since $f(0.5) > 0$ and $f(1.5) < 0$, a root exists in $[0.5, 1.5]$.

2. First midpoint: $c_1 = \frac{0.5 + 1.5}{2} = 1.0, \quad f(1.0) = e^{-1} - 3\log(1.0) = 0.3679 - 0 = 0.3679 > 0.$ Update interval: $[x_a, x_b] = [1.0, 1.5]$.

3. Second midpoint: $c_2 = \frac{1.0 + 1.5}{2} = 1.25, \quad f(1.25) = e^{-1.25} - 3\log(1.25) = 0.2865 - 0.6685 = -0.3820 < 0.$ Update interval: $[x_a, x_b] = [1.0, 1.25]$.

4. Third midpoint: $c_3 = \frac{1.0 + 1.25}{2} = 1.125, \quad f(1.125) = e^{-1.125} - 3\log(1.125) = 0.3247 - 0.3955 = -0.0708 < 0.$ Update interval: $[x_a, x_b] = [1.0, 1.125]$.

5. Fourth midpoint: $c_4 = \frac{1.0 + 1.125}{2} = 1.0625, \quad f(1.0625) = e^{-1.0625} - 3\log(1.0625) = 0.3458 - 0.1814 = 0.1644 > 0.$ Update interval: $[x_a, x_b] = [1.0625, 1.125]$.

6. Fifth midpoint: $c_5 = \frac{1.0625 + 1.125}{2} = 1.09375, \quad f(1.09375) = e^{-1.09375} - 3\log(1.09375) = 0.3351 - 0.2868 = 0.0483 > 0.$ Update interval: $[x_a, x_b] = [1.09375, 1.125]$.

7. Sixth midpoint: $c_6 = \frac{1.09375 + 1.125}{2} = 1.109375, \quad f(1.109375) = e^{-1.109375} - 3\log(1.109375) = 0.3293 - 0.3498 = -0.0205 < 0.$ Update interval: $[x_a, x_b] = [1.09375, 1.109375]$.

Convergence:

After several iterations, the root converges to approximately: c \approx 1.1 \quad \text{(to within tolerance \varepsilon = 0.01)}.

Would you like a detailed step-by-step explanation or a plot of the function and iterations?

Related Questions:

1. How does the Bisection Method ensure convergence for continuous functions?
2. Can the Bisection Method handle multiple roots in a single interval?
3. What are the limitations of the Bisection Method compared to Newton’s method?
4. How does changing $\varepsilon$ affect the accuracy and iterations?
5. How can we visualize the function’s behavior over the given interval?

Tip:

Always verify that the function changes signs over the interval $[x_a, x_b]$ before applying the Bisection Method.