pow()

#include <stdio.h>
int main() {
printf("\nNumber = %.2f", pow(2,3)); // Number = 4.00
printf("\nNumber = %.2f", pow(-2,3)); // Number = -8.00
printf("\nNumber = %.2f", pow(2,-3)); // Number = 0.12
printf("\nNumber = %.2f", pow(-2,-3)); // Number = -0.12
printf("\nNumber = %.2f", pow(0.2,4)); // Number = 0.00
return 0;
}

pow(x,y)

Example with integer as inputs

#include <stdio.h>
#include <math.h>
int main(void){
     int  x,y,out;
     x=2;
     y=3;
     out = pow(x,y);
     printf("%d", out);
     return 0;
}

 8

Example with float variables

pow(num1,num2);

16.977535

Example with negative variables

#include <stdio.h>
#include <math.h>
int main(void){
     float   x,y,out;
     x=2.3;
     y=-3.4;
     out = pow(x,y);
     printf("%f", out);
     return 0;
}

0.058901

Example 1: Using User Input for Dynamic Power Calculation

Enhance the tutorial by allowing users to input their base and exponent values.

#include <stdio.h>
#include <math.h>

int main() {
    double base, exponent, result;
    printf("Enter base: ");
    scanf("%lf", &base);
    printf("Enter exponent: ");
    scanf("%lf", &exponent);

    result = pow(base, exponent);
    printf("%.2lf raised to the power of %.2lf is %.2lf\n", base, exponent, result);

    return 0;
}

Enter base: 5
Enter exponent: 3
5.00 raised to the power of 3.00 is 125.00

Example 2: Working with Negative Exponents

Demonstrate how to handle negative exponents using pow().

#include <stdio.h>
#include <math.h>

int main() {
    double base = 2, exponent = -3, result;
    result = pow(base, exponent);
    printf("%.2lf raised to the power of %.2lf is %.5lf\n", base, exponent, result);
    return 0;
}

2.00 raised to the power of -3.00 is 0.12500

Example 3: Handling Fractional Exponents

Using pow() to calculate roots (e.g., square roots, cube roots) by setting fractional exponents.

#include <stdio.h>
#include <math.h>

int main() {
    double base = 27, exponent = 1.0/3.0, result;
    result = pow(base, exponent);
    printf("The cube root of %.2lf is %.2lf\n", base, result);
    return 0;
}

The cube root of 27.00 is 3.00

Example 4: Integer Power Function Using Loops

Provide an alternative to pow() by implementing a custom power function using loops for integer exponents.

#include <stdio.h>

int power(int base, int exponent) {
    int result = 1;
    for (int i = 0; i < exponent; i++) {
        result *= base;
    }
    return result;
}

int main() {
    int base = 3, exponent = 4;
    int result = power(base, exponent);
    printf("%d raised to the power of %d is %d\n", base, exponent, result);
    return 0;
}

3 raised to the power of 4 is 81