find two powers that have a product of 96, with at least one power greater than 1

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96 = 4^2 x 6^1

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To find two powers that have a product of 96, with at least one power greater than 1, we can use trial and error or algebraic methods.

Method 1: Trial and Error
1. Start by listing a few possible powers that are greater than 1. Let's use powers of 2 since they are easy to work with:
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32

2. Divide 96 by each power and see if the other power gives an integer result. We want the product of the two powers to be 96.
96 / 4 = 24 (for 2^2)
96 / 8 = 12 (for 2^3)
96 / 16 = 6 (for 2^4)
96 / 32 = 3 (for 2^5)

We found two powers that give a product of 96: 2^3 and 2^4. So, the two powers that have a product of 96, with at least one power greater than 1, are 8 and 16.

Method 2: Algebraic Method
1. Start by assuming one of the powers as x and the other as y. We want their product to be 96, so we can write the equation:
x * y = 96

2. Since at least one power should be greater than 1, we can assume x > 1 and y > 1.

3. Now, we can start testing different values of x to find corresponding values of y that satisfy the equation.
Let's try x = 2:
2 * y = 96
y = 96 / 2 = 48
However, with this value of y, we don't have at least one power greater than 1.

Let's try x = 3:
3 * y = 96
y = 96 / 3 = 32
With this value of y, we have at least one power greater than 1 (3), and the product is 96.

Thus, we found two powers that have a product of 96, with at least one power greater than 1: 3 and 32.