The choice that is the closest to the result shown above for 4⁻ˣ + 5 = 3ˣ + 4 is x ≈ 3/8. (Correct choice: A)
How to find an approximate solution of a non-linear equation
Let be a non-linear equation of the form f(x) = 0 and differentiable. Then, we can obtain an approximate solution by Newton-Raphson method, a single step iterative numerical method.
The Newton-Raphson formula is described below:
[tex]x_{i+1} = x_{i}-\frac{f(x_{i})}{f'(x_{i})}[/tex] (1)
Where:
- [tex]f(x_{i})[/tex] - Function evaluated at i-th iteration.
- [tex]f'(x_{i})[/tex] - First derivative evaluated at i-th iteration.
In addition, the equation 4⁻ˣ + 5 = 3ˣ + 4 is equivalent to the equation 12ˣ - 4ˣ - 1 = 0. If we know that f(x) = 12ˣ - 4ˣ - 1 = 0, then approximate solution after three iterations is:
f'(x) = 12ˣ · ln 12 - 4ˣ · ln 4 (2)
Step 1
x₁ = 0.5
f(0.5) ≈ 0.464
f'(0.5) ≈ 5.835
x₂ ≈ 0.420
Step 2
x₂ ≈ 0.420
f(0.420) ≈ 0.050
f'(0.420) ≈ 4.575
x₃ ≈ 0.409
Step 3
x₃ ≈ 0.409
f(0.409) ≈ 6.669 × 10⁻⁵
f'(0.409) ≈ 4.422
x₄ ≈ 0.409
The choice that is the closest to the result shown above for 4⁻ˣ + 5 = 3ˣ + 4 is x ≈ 3/8. (Correct choice: A)
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