find real root of equation x+ log base 10 x = 3.375 up to 5 decimal using newton raphson method
Find real root of equation x + \log_{10} x = 3.375 up to 5 decimal using Newton-Raphson method
Answer:
To find the real root of the equation x + \log_{10} x = 3.375 up to 5 decimal places using the Newton-Raphson method, follow these steps:
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Define the function and its derivative:
Let f(x) = x + \log_{10} x - 3.375.
The derivative of f(x), f'(x), is needed for the Newton-Raphson iteration.We have:
f(x) = x + \log_{10} x - 3.375The derivative:
f'(x) = 1 + \frac{1}{x \ln(10)} -
Initialize and set the iteration formula:
Choose an initial guess ( x_0 ). For the Newton-Raphson method, the iteration formula is:x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}We can start with an initial guess. Let’s try ( x_0 = 3 ).
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Execute the Newton-Raphson iterations:
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First iteration:
f(3) = 3 + \log_{10} 3 - 3.375 \approx -0.225309f'(3) = 1 + \frac{1}{3 \ln(10)} \approx 1.144716x_1 = 3 - \frac{-0.225309}{1.144716} \approx 3.196801 -
Second iteration:
f(3.196801) = 3.196801 + \log_{10} 3.196801 - 3.375 \approx 0.025349f'(3.196801) = 1 + \frac{1}{3.196801 \ln(10)} \approx 1.135920x_2 = 3.196801 - \frac{0.025349}{1.135920} \approx 3.174496 -
Third iteration:
f(3.174496) = 3.174496 + \log_{10} 3.174496 - 3.375 \approx 0.000123f'(3.174496) = 1 + \frac{1}{3.174496 \ln(10)} \approx 1.136978x_3 = 3.174496 - \frac{0.000123}{1.136978} \approx 3.174388 -
Fourth iteration (for higher precision):
f(3.174388) = 3.174388 + \log_{10} 3.174388 - 3.375 \approx 9.7 \times 10^{-8}f'(3.174388) = 1 + \frac{1}{3.174388 \ln(10)} \approx 1.136982x_4 = 3.174388 - \frac{9.7 \times 10^{-8}}{1.136982} \approx 3.174388
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Verify the final result:
We reach ( x = 3.174388 ) within the precision limit.
Final Answer:
The real root of the equation x + \log_{10} x = 3.375 up to 5 decimal places using the Newton-Raphson method is approximately ( \boxed{3.17439} ).