If d is the highest common factor (HCF) of 56 and 72, how can we find the values of x and y that satisfy the equation d * 56^x * 72^y, and why are x and y not unique?

To solve this problem, we need to first determine the highest common factor (HCF) of the numbers 56 and 72.

The prime factorization of 56 is:

• 56 = 2^3 * 7^1

Now, for 72, the prime factorization is:

• 72 = 2^3 * 3^2

To find the HCF, we take the lowest power of each common prime factor:

• The only prime factor common to both numbers is 2, and the lowest power of 2 in both factorizations is 2^3.

Therefore, the HCF (d) of 56 and 72 is:

• d = 2^3 = 8

Now that we have found d, we can express the equation: d * 56^x * 72^y.

Substituting our value of d into the equation, we have:

8 * 56^x * 72^y.

We can further break down this equation using the prime factorizations:

Substituting the factorizations:

8 * (2^3 * 7^1)^x * (2^3 * 3^2)^y.

This becomes:

(2^3)¹ * (2^3)^x * (2^3)^y * (7^1)^x * (3^2)^y.

Combining the terms with common bases:

2^{3 + 3x + 3y} * 7^x * 3^2y.

Now we need to find values for x and y that maintain the equality of the initial structured expression. Since there are no restrictions on x and y, we can see that there are infinitely many pairs (x,y) that can satisfy this equation. For example:

• If x = 0 and y = 0, the equation simplifies to 8.
• If x = 1 and y = 0, it becomes 8 * 56 = 448.
• If x = 0 and y = 1, it becomes 8 * 72 = 576.
• If x = 1 and y = 1, it becomes 8 * 56 * 72 = 32256.

Thus, we can conclude that x and y are not unique since any combination of x and y that satisfies the equation will work as long as the structure of the base number and exponent remains consistent. Therefore, we have shown that there are multiple (in fact, infinitely many) pairs (x, y) that fit this scenario.