Question

How to use log2() function to calculate binary base - 2 logarithm in C

Answer 1

#include <stdio.h>     
#include <math.h> 
 
int main(int argc, char **argv)
{   
    printf("log2(1) = %f\n", log2(1));    
	printf("log2(1024) = %f\n", log2(1024));
	printf("log2(512) = %f\n", log2(512));
	printf("log2(5) = %f\n", log2(5));
	printf("log2(100) = %f\n", log2(100));
	printf("log2(125) = %f\n", log2(125));	
	printf("log2(125) / log2(5) = %f\n", log2(125) / log2(5));
	printf("log2(3.5) = %f\n", log2(3.5));
	printf("log2(1000) = %f\n", log2(1000));
    printf("log2(0.001) = %f\n", log2(0.001));
  
    return 0;
}
 
 
/*
run:
   
log2(1) = 0.000000
log2(1024) = 10.000000
log2(512) = 9.000000
log2(5) = 2.321928
log2(100) = 6.643856
log2(125) = 6.965784
log2(125) / log2(5) = 3.000000
log2(3.5) = 1.807355
log2(1000) = 9.965784
log2(0.001) = -9.965784

*/

 

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