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),
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
… [Trackback]
[…] Find More on that Topic: geeksforgeeks.org/find-the-ln-x-and-log10x-with-the-help-of-expansion/ […]