Question

How to use log1p() function to calculate natural base e logarithm of 1 + argument in C++

Answer 1

#include <iostream>
#include <cmath>

using namespace std;

int main()
{
	cout << "log1p(1) = " << log1p(1) << endl;
	cout << "log1p(5) = " << log1p(5) << endl;
	cout << "log1p(0) = " << log1p(0) << endl;
	cout << "log1p(-0) = " << log1p(-0) << endl;
	cout << "log1p(100) = " << log1p(100) << endl;
	cout << "log1p(125) = " << log1p(125) << endl;
	cout << "log1p(125) / log1p(5) = " << log1p(125) / log1p(5) << endl;
	cout << "log1p(3.5) = " << log1p(3.5) << endl;
	cout << "log1p(1000) = " << log1p(1000) << endl;
	cout << "log1p(0.001) = " << log1p(0.001) << endl;

	return 0;
}

/*
run:

log1p(1) = 0.693147
log1p(5) = 1.79176
log1p(0) = 0
log1p(-0) = 0
log1p(100) = 4.61512
log1p(125) = 4.83628
log1p(125) / log1p(5) = 2.69918
log1p(3.5) = 1.50408
log1p(1000) = 6.90875
log1p(0.001) = 0.0009995

*/

 

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