Given a positive number x, the task is to find the natural log (ln) and log to the base 10 (log10) of this number with the help of expansion.
Example:
Input: x = 5
Output: ln 5.000 = 1.609
log10 5.000 = 0.699
Input: x = 10
Output: ln 10.000 = 2.303
log10 10.000 = 1.000
Approach:
- The expansion of natural logarithm of x (ln x) is:
- Therefore this series can be summed up as:
- Hence a function can be made to evaluate the nth term of the sequence for 1 ? x ? n
- Now to calculate log10 x, below formula can be used:
Below is the implementation of the above approach:
C++
// CPP code to Find the ln x and// log<sub>10</sub> x with the help of expansion#include <cmath>#include <iomanip>#include <iostream>using namespace std;// Function to calculate ln x using expansiondouble calculateLnx(double n){ double num, mul, cal, sum = 0; num = (n - 1) / (n + 1); // terminating value of the loop // can be increased to improve the precision for (int i = 1; i <= 1000; i++) { mul = (2 * i) - 1; cal = pow(num, mul); cal = cal / mul; sum = sum + cal; } sum = 2 * sum; return sum;}// Function to calculate log10 xdouble calculateLogx(double lnx){ return (lnx / 2.303);}// Driver Codeint main(){ double lnx, logx, n = 5; lnx = calculateLnx(n); logx = calculateLogx(lnx); // setprecision(3) is used to display // the output up to 3 decimal places cout << fixed << setprecision(3) << "ln " << n << " = " << lnx << endl; cout << fixed << setprecision(3) << "log10 " << n << " = " << logx << endl;} |
Java
// Java code to Find the ln x and// log<sub>10</sub> x with the help of expansionimport java.io.*;class GFG { // Function to calculate ln x using expansionstatic double calculateLnx(double n){ double num, mul, cal, sum = 0; num = (n - 1) / (n + 1); // terminating value of the loop // can be increased to improve the precision for (int i = 1; i <= 1000; i++) { mul = (2 * i) - 1; cal = Math.pow(num, mul); cal = cal / mul; sum = sum + cal; } sum = 2 * sum; return sum;}// Function to calculate log10 xstatic double calculateLogx(double lnx){ return (lnx / 2.303);}// Driver Codepublic static void main (String[] args) { double lnx, logx, n = 5; lnx = calculateLnx(n); logx = calculateLogx(lnx); // setprecision(3) is used to display // the output up to 3 decimal places System.out.println ("ln " + n + " = " + lnx ); System.out.println ("log10 " + n + " = "+ logx );}}// This code is contributed by ajit |
Python3
# Python 3 code to Find the ln x and# log<sub>10</sub> x with the help of expansion# Function to calculate ln x using expansionfrom math import powdef calculateLnx(n): sum = 0 num = (n - 1) / (n + 1) # terminating value of the loop # can be increased to improve the precision for i in range(1, 1001, 1): mul = (2 * i) - 1 cal = pow(num, mul) cal = cal / mul sum = sum + cal sum = 2 * sum return sum# Function to calculate log10 xdef calculateLogx(lnx): return (lnx / 2.303)# Driver Codeif __name__ == '__main__': n = 5 lnx = calculateLnx(n) logx = calculateLogx(lnx) # setprecision(3) is used to display # the output up to 3 decimal places print("ln", "{0:.3f}".format(n), "=", "{0:.3f}".format(lnx)) print("log10", "{0:.3f}".format(n), "=", "{0:.3f}".format(logx)) # This code is contributed by# Surendra_Gangwar |
C#
// C# code to Find the ln x and// log<sub>10</sub> x with the help of expansionusing System; class GFG { // Function to calculate ln x using expansionstatic double calculateLnx(double n){ double num, mul, cal, sum = 0; num = (n - 1) / (n + 1); // terminating value of the loop // can be increased to improve the precision for (int i = 1; i <= 1000; i++) { mul = (2 * i) - 1; cal = Math.Pow(num, mul); cal = cal / mul; sum = sum + cal; } sum = 2 * sum; return sum;}// Function to calculate log10 xstatic double calculateLogx(double lnx){ return (lnx / 2.303);}// Driver Codepublic static void Main (String[] args) { double lnx, logx, n = 5; lnx = calculateLnx(n); logx = calculateLogx(lnx); // setprecision(3) is used to display // the output up to 3 decimal places Console.WriteLine("ln " + n + " = " + lnx ); Console.WriteLine("log10 " + n + " = "+ logx );}}// This code is contributed by Princi Singh |
Javascript
<script>// Javascript code to Find the ln x and// log<sub>10</sub> x with the help of expansion// Function to calculate ln x using expansionfunction calculateLnx(n){ let num, mul, cal, sum = 0; num = (n - 1) / (n + 1); // Terminating value of the loop // can be increased to improve the precision for(let i = 1; i <= 1000; i++) { mul = (2 * i) - 1; cal = Math.pow(num, mul); cal = cal / mul; sum = sum + cal; } sum = 2 * sum; return sum;}// Function to calculate log10 xfunction calculateLogx(lnx){ return (lnx / 2.303);}// Driver Codelet lnx, logx, n = 5;lnx = calculateLnx(n);logx = calculateLogx(lnx);// setprecision(3) is used to display// the output up to 3 decimal placesdocument.write("ln " + n + " = " + lnx + "<br>");document.write("log10 " + n + " = "+ logx + "<br>");// This code is contributed by souravmahato348</script> |
Output
ln 5.000 = 1.609 log10 5.000 = 0.699
Time complexity: O(1000), as the loop iterates 1000 times.
Auxiliary Space: O(1),
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