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

1 Answer

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#include <stdio.h>     
#include <math.h> 
 
int main(int argc, char **argv)
{   
    printf("log1p(1) = %f\n", log1p(1));    
	printf("log1p(0) = %f\n", log1p(0));
	printf("log1p(-0) = %f\n", log1p(-0));
	printf("log1p(5) = %f\n", log1p(5));
	printf("log1p(100) = %f\n", log1p(100));
	printf("log1p(125) = %f\n", log1p(125));	
	printf("log1p(125) / log1p(5) = %f\n", log1p(125) / log1p(5));
	printf("log1p(3.5) = %f\n", log1p(3.5));
	printf("log1p(1000) = %f\n", log1p(1000));
    printf("log1p(0.001) = %f\n", log1p(0.001));
  
    return 0;
}
 
 
/*
run:
   
log1p(1) = 0.693147
log1p(0) = 0.000000
log1p(-0) = 0.000000
log1p(5) = 1.791759
log1p(100) = 4.615121
log1p(125) = 4.836282
log1p(125) / log1p(5) = 2.699180
log1p(3.5) = 1.504077
log1p(1000) = 6.908755
log1p(0.001) = 0.001000

*/

 



answered Mar 26, 2016 by avibootz
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