Question

6. Evaluate
2) - 2
- 2
(a)
3
of the​

Answer 1

Answer:

You would solve (simplify) the numerical expression 2^(5/2) ‒ 2^(3/2) by using the properties of exponents.

First, we start by utilizing the definition governing rational exponents:

“If m/n is a rational number, where n is a positive integer greater than one, and b is a real number such that b^(1/n) is a real number and not imaginary, then b^(m/n) = [b^m]^(1/n) = [b^(1/n)]^m.”

Therefore, we can write:

1.) 2^(5/2) ‒ 2^(3/2) = [2^½]^5 – [2^½]^3 , where b = 2, 1/n = ½, and m

= 5 and 3. Now, we have …

2.) = (2^½)(2^½)(2^½)(2^½)(2^½) ‒ (2^½)(2^½)(2^½) by a property of exponents: b^n = (b)(b)(b)(b) … (b) for n factors of b, where b is any real number and n is a positive integer. In this case, b = 2^½ and n = 5 and 3.

3.) = [(2^½)^4](2^½) ‒ [(2^½)^2](2^½). Since equality is symmetrical, i.e., if a = b, then b = a, then (b)(b)(b) … (b) = b^n for n factors of b, where b is any real number and n is a positive integer. In this case, b = 2^½ and n = 4 and 2.

4.) = [2^(4/2)](2^½) ‒ [2^(2/2)](2^½) because [b^(1/n)]^m = b^(m/n) since equality is symmetrical.

5.) = [2^(2)](2^½) ‒ [2^(1)](2^½)

6.) = (4)[2^(½)] ‒ (2)[2^(½)] since, in the 2nd term, a^1 = a for any real number “a”.

7.) = (4 ‒ 2)[2^(½)] by factoring out 2^(½); therefore, we have:

8.) 2^(5/2) ‒ 2^(3/2) = 2[2^(1/2)]

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