How to calculate exponent (2^2x)(4^x)(8^-x)?
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:
When you have powers of powers, you multiply the exponents, so we can rewrite it as:
And since we now have the same base, we can add the exponents.
The exponent on the final answer is 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...
According to laws of exponents, when you multiply powers, the exponents are added. Therefore...
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 * 2^2x * 2^-3x
As the base is the same, we can add the exponents
=> 2^(2x + 2x - 3x)
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:
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.