Skip to content
Related Articles

Related Articles

log() function in C++

View Discussion
Improve Article
Save Article
  • Difficulty Level : Basic
  • Last Updated : 18 Aug, 2022
View Discussion
Improve Article
Save Article

Log() function in C++ : The log() function in C++ returns the natural logarithm (base-e logarithm) of the argument passed in the parameter.

logarithmic-function 

Syntax for returning natural logarithm: result = log(x) 

Syntax for returning logarithm (base-10 logarithm) of the argument. result = log10(x)

The parameters can be of any data type like int, double or float or long double. Log() function returns value according to the following conditions are as follows:

a) if x>1 then positive ..
b) if 0<x<1 returns a negative value ..
c) if x=1 then it returns 0 ..
d) if x=0 then it returns -inf ..
e) if x<0 then it returns NaN(not a number) 

CPP




// CPP program to implement log() function
#include <bits/stdc++.h>
using namespace std;
 
// function to evaluate natural logarithm base-e
double valueE(double d)
{
    return log(d);
}
 
// function to evaluate logarithm base-10
double value10(double d)
{
    return log10(d);
}
 
// driver program to test the above function
int main()
{
    double d = 10;
    cout << "The logarithm value(base-e) of " << d
         << " is " << valueE(d) << endl;
    cout << "The logarithm value(base-10) of " << d
         << " is " << value10(d) << endl;
    return 0;
}

Output:

The logarithm value(base-e) of 10 is 2.30259
The logarithm value(base-10) of 10 is 1

Time Complexity: O(1)

Auxiliary Space: O(1)

Application: One of the applications of log() function is to calculate values related to log, for e.g. while finding polite number we need the formula to be written in code, for that we can use log() function. Given below is an implementation of log() function. 

CPP




// CPP program to find Nth polite number
 
#include& lt; bits / stdc++.h & gt;
 
using namespace std;
 
// function to evaluate n-th polite number
double polite(double n)
{
    n += 1;
    double base = 2;
    return n
           + (log((n + (log(n) / log(base))))) / log(base);
}
 
// driver code
int main()
{
    double n = 7;
    cout& lt;
    <
    (int)polite(n);
    return 0;
}

Output:

11 

Time Complexity: O(1)

Auxiliary Space: O(1)


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!