^ in java does not mean to raise to a power. It means XOR.

You can use java's Math.pow()


And you might want to consider using double instead of intโ€”that is:

double height;
double weight;

Note that 199/100 evaluates to 1.

Answer from WuHoUnited on Stack Overflow
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GeeksforGeeks
geeksforgeeks.org โ€บ java โ€บ math-pow-method-in-java-with-example
Math pow() Method in Java with Example - GeeksforGeeks
March 28, 2025 - Example 1: This example demonstrates how to use the Math.pow() method in Java to calculate the power of a number (base raised to the exponent).
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Oracle
docs.oracle.com โ€บ javase โ€บ 8 โ€บ docs โ€บ api โ€บ java โ€บ lang โ€บ Math.html
Math (Java Platform SE 8 )
October 20, 2025 - if the second argument is a finite odd integer, the result is equal to the negative of the result of raising the absolute value of the first argument to the power of the second argument ยท if the second argument is finite and not an integer, then the result is NaN. If both arguments are integers, then the result is exactly equal to the mathematical result of raising the first argument to the power of the second argument if that result can in fact be represented exactly as a double value.
Discussions

Raising a number to a power in Java - Stack Overflow
Here is my code. For some reason my BMI is not calculated correctly. When I check the output on a calculator for this : (10/((10/100)^2))) I get 1000, but in my program, I get 5. I'm not sure wha... More on stackoverflow.com
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What's the algorithm behind Math.pow() in Java - Stack Overflow
I just started with Java and as a first project I am writing a program that finds roots(cube root in this case) of a given number. Presently I am trying out Newton-Ralphson to achieve this. Here is... More on stackoverflow.com
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Explain Java function Math.pow( ) with an example.
In Java, Math.pow( ) function returns the value of the first argument raised to the power of the second argument. More on knowledgeboat.com
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January 16, 2023
Can somebody please help me with Math.pow and specifically raising to the power of.
This is what I have at the minute but it doesn't work. In what way doesn't it work? Please let us know the actual error you're getting. For your pasted code, Java syntax is much more rigid than math syntax. See here for a reference on Java operators. In particular, in math notation, multiplication is implied when you put two things next to each other. That's not true of Java. You need to explicitly use the multiplication operator. EDIT: It also seems you need to define k and ans unless they're instance variables with definitions you haven't included with your code snippet. If you haven't programmed in Java I'd strongly recommend reading some reference material. There are docs on the official Oracle site here More on reddit.com
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June 21, 2013
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W3Schools
w3schools.com โ€บ java โ€บ ref_math_pow.asp
Java Math pow() Method
System.out.println(Math.pow(2, ...ln(Math.pow(8, -1)); System.out.println(Math.pow(10, -2)); ... The pow() method raises a number to the power of another number....
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W3Schools
w3schools.com โ€บ jsref โ€บ jsref_pow.asp
W3Schools.com
The Math.pow() method returns the value of x to the power of y (xy). let a = Math.pow(0, 1); let b = Math.pow(1, 1); let c = Math.pow(1, 10); let d = Math.pow(3, 3); let e = Math.pow(-3, 3); let f = Math.pow(2, 4); Try it Yourself ยป ยท Math.pow() ...
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Codecademy
codecademy.com โ€บ docs โ€บ java โ€บ math methods โ€บ .pow()
Java | Math Methods | .pow() | Codecademy
June 23, 2025 - The Math.pow() method in Java is a static method that calculates the value of a base number raised to the power of an exponent. It performs exponentiation operations by taking two double parameters and returning the result as a double value.
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Programiz
programiz.com โ€บ java-programming โ€บ examples โ€บ power-number
Java Program to Calculate the Power of a Number
In this program, we use Java's Math.pow() function to calculate the power of the given base.
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This is a fun question. If you look into the the source code for Java's Math class, you will find that it calls StrictMath.pow(double1, double2), and StrictMath's signature is public static native double pow(double a, double b);

So, in the end, it is a truly native call that might differ depending on the platform. However, there is an implementation somewhere, and it isn't very easy to look at. Here is the description of the function and the code for the function itself:

Note

Looking through the math, trying to understand it might inevitably lead to even more questions. But, by searching through this Github on Java Math Function Source Code and glancing out the mathematical summaries, you can definitely understand the native functions better. Happy Exploring :)

Method Description

Method:  Let x =  2   * (1+f)
      1. Compute and return log2(x) in two pieces:
              log2(x) = w1 + w2,
         where w1 has 53-24 = 29 bit trailing zeros.
      2. Perform y*log2(x) = n+y' by simulating muti-precision
         arithmetic, where |y'|<=0.5.
      3. Return x**y = 2**n*exp(y'*log2)

Special Cases

      1.  (anything) ** 0  is 1
      2.  (anything) ** 1  is itself
      3.  (anything) ** NAN is NAN
      4.  NAN ** (anything except 0) is NAN
      5.  +-(|x| > 1) **  +INF is +INF
      6.  +-(|x| > 1) **  -INF is +0
      7.  +-(|x| < 1) **  +INF is +0
      8.  +-(|x| < 1) **  -INF is +INF
      9.  +-1         ** +-INF is NAN
      10. +0 ** (+anything except 0, NAN)               is +0
      11. -0 ** (+anything except 0, NAN, odd integer)  is +0
      12. +0 ** (-anything except 0, NAN)               is +INF
      13. -0 ** (-anything except 0, NAN, odd integer)  is +INF
      14. -0 ** (odd integer) = -( +0 ** (odd integer) )
      15. +INF ** (+anything except 0,NAN) is +INF
      16. +INF ** (-anything except 0,NAN) is +0
      17. -INF ** (anything)  = -0 ** (-anything)
      18. (-anything) ** (integer) is (-1)**(integer)*(+anything**integer)
      19. (-anything except 0 and inf) ** (non-integer) is NAN

Accuracy

       pow(x,y) returns x**y nearly rounded. In particular
                      pow(integer,integer)
       always returns the correct integer provided it is
       representable.

Source Code

#ifdef __STDC__
        double __ieee754_pow(double x, double y)
#else
        double __ieee754_pow(x,y)
        double x, y;
#endif
{
        double z,ax,z_h,z_l,p_h,p_l;
        double y1,t1,t2,r,s,t,u,v,w;
        int i0,i1,i,j,k,yisint,n;
        int hx,hy,ix,iy;
        unsigned lx,ly;

        i0 = ((*(int*)&one)>>29)^1; i1=1-i0;
        hx = __HI(x); lx = __LO(x);
        hy = __HI(y); ly = __LO(y);
        ix = hx&0x7fffffff;  iy = hy&0x7fffffff;

    /* y==zero: x**0 = 1 */
        if((iy|ly)==0) return one;

    /* +-NaN return x+y */
        if(ix > 0x7ff00000 || ((ix==0x7ff00000)&&(lx!=0)) ||
           iy > 0x7ff00000 || ((iy==0x7ff00000)&&(ly!=0)))
                return x+y;

    /* determine if y is an odd int when x < 0
     * yisint = 0       ... y is not an integer
     * yisint = 1       ... y is an odd int
     * yisint = 2       ... y is an even int
     */
        yisint  = 0;
        if(hx<0) {
            if(iy>=0x43400000) yisint = 2; /* even integer y */
            else if(iy>=0x3ff00000) {
                k = (iy>>20)-0x3ff;        /* exponent */
                if(k>20) {
                    j = ly>>(52-k);
                    if((j<<(52-k))==ly) yisint = 2-(j&1);
                } else if(ly==0) {
                    j = iy>>(20-k);
                    if((j<<(20-k))==iy) yisint = 2-(j&1);
                }
            }
        }

    /* special value of y */
        if(ly==0) {
            if (iy==0x7ff00000) {       /* y is +-inf */
                if(((ix-0x3ff00000)|lx)==0)
                    return  y - y;      /* inf**+-1 is NaN */
                else if (ix >= 0x3ff00000)/* (|x|>1)**+-inf = inf,0 */
                    return (hy>=0)? y: zero;
                else                    /* (|x|<1)**-,+inf = inf,0 */
                    return (hy<0)?-y: zero;
            }
            if(iy==0x3ff00000) {        /* y is  +-1 */
                if(hy<0) return one/x; else return x;
            }
            if(hy==0x40000000) return x*x; /* y is  2 */
            if(hy==0x3fe00000) {        /* y is  0.5 */
                if(hx>=0)       /* x >= +0 */
                return sqrt(x);
            }
        }

        ax   = fabs(x);
    /* special value of x */
        if(lx==0) {
            if(ix==0x7ff00000||ix==0||ix==0x3ff00000){
                z = ax;                 /*x is +-0,+-inf,+-1*/
                if(hy<0) z = one/z;     /* z = (1/|x|) */
                if(hx<0) {
                    if(((ix-0x3ff00000)|yisint)==0) {
                        z = (z-z)/(z-z); /* (-1)**non-int is NaN */
                    } else if(yisint==1)
                        z = -1.0*z;             /* (x<0)**odd = -(|x|**odd) */
                }
                return z;
            }
        }

        n = (hx>>31)+1;

    /* (x<0)**(non-int) is NaN */
        if((n|yisint)==0) return (x-x)/(x-x);

        s = one; /* s (sign of result -ve**odd) = -1 else = 1 */
        if((n|(yisint-1))==0) s = -one;/* (-ve)**(odd int) */

    /* |y| is huge */
        if(iy>0x41e00000) { /* if |y| > 2**31 */
            if(iy>0x43f00000){  /* if |y| > 2**64, must o/uflow */
                if(ix<=0x3fefffff) return (hy<0)? huge*huge:tiny*tiny;
                if(ix>=0x3ff00000) return (hy>0)? huge*huge:tiny*tiny;
            }
        /* over/underflow if x is not close to one */
            if(ix<0x3fefffff) return (hy<0)? s*huge*huge:s*tiny*tiny;
            if(ix>0x3ff00000) return (hy>0)? s*huge*huge:s*tiny*tiny;
        /* now |1-x| is tiny <= 2**-20, suffice to compute
           log(x) by x-x^2/2+x^3/3-x^4/4 */
            t = ax-one;         /* t has 20 trailing zeros */
            w = (t*t)*(0.5-t*(0.3333333333333333333333-t*0.25));
            u = ivln2_h*t;      /* ivln2_h has 21 sig. bits */
            v = t*ivln2_l-w*ivln2;
            t1 = u+v;
            __LO(t1) = 0;
            t2 = v-(t1-u);
        } else {
            double ss,s2,s_h,s_l,t_h,t_l;
            n = 0;
        /* take care subnormal number */
            if(ix<0x00100000)
                {ax *= two53; n -= 53; ix = __HI(ax); }
            n  += ((ix)>>20)-0x3ff;
            j  = ix&0x000fffff;
        /* determine interval */
            ix = j|0x3ff00000;          /* normalize ix */
            if(j<=0x3988E) k=0;         /* |x|<sqrt(3/2) */
            else if(j<0xBB67A) k=1;     /* |x|<sqrt(3)   */
            else {k=0;n+=1;ix -= 0x00100000;}
            __HI(ax) = ix;

        /* compute ss = s_h+s_l = (x-1)/(x+1) or (x-1.5)/(x+1.5) */
            u = ax-bp[k];               /* bp[0]=1.0, bp[1]=1.5 */
            v = one/(ax+bp[k]);
            ss = u*v;
            s_h = ss;
            __LO(s_h) = 0;
        /* t_h=ax+bp[k] High */
            t_h = zero;
            __HI(t_h)=((ix>>1)|0x20000000)+0x00080000+(k<<18);
            t_l = ax - (t_h-bp[k]);
            s_l = v*((u-s_h*t_h)-s_h*t_l);
        /* compute log(ax) */
            s2 = ss*ss;
            r = s2*s2*(L1+s2*(L2+s2*(L3+s2*(L4+s2*(L5+s2*L6)))));
            r += s_l*(s_h+ss);
            s2  = s_h*s_h;
            t_h = 3.0+s2+r;
            __LO(t_h) = 0;
            t_l = r-((t_h-3.0)-s2);
        /* u+v = ss*(1+...) */
            u = s_h*t_h;
            v = s_l*t_h+t_l*ss;
        /* 2/(3log2)*(ss+...) */
            p_h = u+v;
            __LO(p_h) = 0;
            p_l = v-(p_h-u);
            z_h = cp_h*p_h;             /* cp_h+cp_l = 2/(3*log2) */
            z_l = cp_l*p_h+p_l*cp+dp_l[k];
        /* log2(ax) = (ss+..)*2/(3*log2) = n + dp_h + z_h + z_l */
            t = (double)n;
            t1 = (((z_h+z_l)+dp_h[k])+t);
            __LO(t1) = 0;
            t2 = z_l-(((t1-t)-dp_h[k])-z_h);
        }

    /* split up y into y1+y2 and compute (y1+y2)*(t1+t2) */
        y1  = y;
        __LO(y1) = 0;
        p_l = (y-y1)*t1+y*t2;
        p_h = y1*t1;
        z = p_l+p_h;
        j = __HI(z);
        i = __LO(z);
        if (j>=0x40900000) {                            /* z >= 1024 */
            if(((j-0x40900000)|i)!=0)                   /* if z > 1024 */
                return s*huge*huge;                     /* overflow */
            else {
                if(p_l+ovt>z-p_h) return s*huge*huge;   /* overflow */
            }
        } else if((j&0x7fffffff)>=0x4090cc00 ) {        /* z <= -1075 */
            if(((j-0xc090cc00)|i)!=0)           /* z < -1075 */
                return s*tiny*tiny;             /* underflow */
            else {
                if(p_l<=z-p_h) return s*tiny*tiny;      /* underflow */
            }
        }
    /*
     * compute 2**(p_h+p_l)
     */
        i = j&0x7fffffff;
        k = (i>>20)-0x3ff;
        n = 0;
        if(i>0x3fe00000) {              /* if |z| > 0.5, set n = [z+0.5] */
            n = j+(0x00100000>>(k+1));
            k = ((n&0x7fffffff)>>20)-0x3ff;     /* new k for n */
            t = zero;
            __HI(t) = (n&~(0x000fffff>>k));
            n = ((n&0x000fffff)|0x00100000)>>(20-k);
            if(j<0) n = -n;
            p_h -= t;
        }
        t = p_l+p_h;
        __LO(t) = 0;
        u = t*lg2_h;
        v = (p_l-(t-p_h))*lg2+t*lg2_l;
        z = u+v;
        w = v-(z-u);
        t  = z*z;
        t1  = z - t*(P1+t*(P2+t*(P3+t*(P4+t*P5))));
        r  = (z*t1)/(t1-two)-(w+z*w);
        z  = one-(r-z);
        j  = __HI(z);
        j += (n<<20);
        if((j>>20)<=0) z = scalbn(z,n); /* subnormal output */
        else __HI(z) += (n<<20);
        return s*z;
}
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CodeGym
codegym.cc โ€บ java blog โ€บ java math โ€บ math pow() method in java
Math.pow() Method in Java
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Medium
medium.com โ€บ @AlexanderObregon โ€บ javas-math-pow-method-explained-7c0f746ad420
Understanding Java's Math.pow() Method | Medium
July 16, 2024 - Javaโ€™s Math.pow() method is a powerful tool for performing exponentiation, allowing developers to raise a number to a specified power.
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educative.io โ€บ answers โ€บ how-to-use-the-mathpow-method-in-java
How to use the Math.pow() method in Java
The Math.pow() is an built-in method in Java Math class and is used to calculate the power of a given number.
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LeetCode
leetcode.com โ€บ problems โ€บ powx-n
Pow(x, n) - LeetCode
Can you solve this real interview question? Pow(x, n) - Implement pow(x, n) [http://www.cplusplus.com/reference/valarray/pow/], which calculates x raised to the power n (i.e., xn).
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carmatec.com โ€บ home โ€บ java math.pow() explained: the java power function
Java Math.pow() Explained: The Java Power Function
January 7, 2026 - A negative base with a non-integer exponent typically returns NaN (e.g., Math.pow(-4, 0.5) โ†’ square root of negative). Bases of magnitude greater than 1 raised to +Infinity return +Infinity. Bases between 0 and 1 raised to +Infinity return +0.0. Overflow results in +Infinity or -Infinity; underflow results in +0.0 or -0.0. ... java System.out.println(Math.pow(Double.NaN, 5)); // NaN System.out.println(Math.pow(-4, 0.5)); // NaN System.out.println(Math.pow(0, 0)); // 1.0 System.out.println(Math.pow(2, Double.POSITIVE_INFINITY)); // Infinity
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Baeldung
baeldung.com โ€บ home โ€บ java โ€บ java numbers โ€บ using math.pow in java
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Vultr Docs
docs.vultr.com โ€บ java โ€บ standard-library โ€บ java โ€บ lang โ€บ Math โ€บ pow
Java Math pow() - Calculate Power
December 3, 2024 - This code snippet computes (2^3), resulting in 8.0. Here, Math.pow() takes two double arguments and returns the base raised to the power of the exponent as a double. Use integer values that Java implicitly converts to double.
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Tutorialspoint
tutorialspoint.com โ€บ java โ€บ number_pow.htm
Java Math.pow() Method
This method returns the value of the first argument raised to the power of the second argument. public class Test { public static void main(String args[]) { double x = 11.635; double y = 2.76; System.out.printf("The value of e is %.4f%n", Math.E); ...