We need to convert each term to the same base, in this case, 2.

The first term already has a base of 2. The second term has a base of 4, which is 2^2, and the third term has a base of 8, which is the same as 2^3. So we have:

[2^(2x)][4^x][8^(-x)]

= [2^(2x)][(2^2)^x][(2^3)^(-x)]

When you have powers of powers, you multiply the exponents, so we can rewrite it as:

= [2^2x][2^2x][(2^(-3x)]

And since we now have the same base, we can add the exponents.

= 2^(2x+2x-3x)

= 2^x

**The exponent on the final answer is x.**

(2^2x)(4^x)(8^-x)

To multiply powers, the powers must have common bases. In this example, the common base is 2.

4 = 2^2 Therefore... 4^x = 2^2x

8 = 2^3 Therefore... 8^-x = 2^-3x

With common bases, the problem becomes...

(2^2x)(2^2x)(2^-3x)

According to laws of exponents, when you multiply powers, the exponents are added. Therefore...

(2^2x)(2^2x)(2^-3x)

2^(2x + 2x + -3x)

2^1x or 2^x

**Simplified answer: 2^x**

**The exponent of the simplified answer is x.**

We have to find the exponent of (2^2x)(4^x)(8^-x)

(2^2x)(4^x)(8^-x)

=> 2^2x * 2^2x * 2^-3x

As the base is the same, we can add the exponents

=> 2^(2x + 2x - 3x)

=> 2^x

**The exponent is x.**

First, we'll create matching bases to all factors.

4^x = (2^2)^x = 2^2x

8^(-x) = 2^(-3x)

Now, we'll perform the multiplication:

(2^2x)*(2^2x)*(2^-3x)

Since the bases are matching, we can add the superscripts:

(2^2x)*(2^2x)*(2^-3x) = 2^(2x + 2x - 3x) = 2^x

**The requested exponent of 2^x is x.**