# Find the ln(X) and log10X with the help of expansion

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


## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

1. The expansion of natural logarithm of x (ln x) is:

2. Therefore this series can be summed up as: 3. Hence a function can be made to evaluate the nth term of the sequence for 1 &leq; x &leq; n
4. 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  // log10 x with the help of expansion     #include  #include  #include     using namespace std;     // Function to calculate ln x using expansion  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 = pow(num, mul);          cal = cal / mul;          sum = sum + cal;      }      sum = 2 * sum;      return sum;  }     // Function to calculate log10 x  double calculateLogx(double lnx)  {      return (lnx / 2.303);  }     // Driver Code  int 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  // log10 x with the help of expansion  import java.io.*;     class GFG   {         // Function to calculate ln x using expansion  static 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 x  static double calculateLogx(double lnx)  {      return (lnx / 2.303);  }     // Driver Code  public 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  # log10 x with the help of expansion  # Function to calculate ln x using expansion  from math import pow def 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 x  def calculateLogx(lnx):      return (lnx / 2.303)     # Driver Code  if __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  // log10 x with the help of expansion  using System;         class GFG   {         // Function to calculate ln x using expansion  static 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 x  static double calculateLogx(double lnx)  {      return (lnx / 2.303);  }     // Driver Code  public 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

Output:

ln 5.000 = 1.609
log10 5.000 = 0.699


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