Certainly! Let's explore the idea of extending the Product Property of Exponents to a fractional exponent like [tex]\(\frac{1}{2}\)[/tex] step by step.
Firstly, let's recall the Product Property of Exponents:
[tex]\[ a^{m} \cdot a^{n} = a^{m+n} \][/tex]
When extending to a fractional exponent, let's choose [tex]\(a = 9\)[/tex] and the fractional exponent [tex]\( \frac{1}{2} \)[/tex].
### Part (a)
Step 1: Calculate [tex]\( 9^{\frac{1}{2}} \)[/tex].
The exponent [tex]\(\frac{1}{2}\)[/tex] signifies the square root. So,
[tex]\[ 9^{\frac{1}{2}} = \sqrt{9} \][/tex]
The square root of 9 is 3 (since [tex]\(3^2 = 9\)[/tex]):
[tex]\[ \sqrt{9} = 3 \][/tex]
Thus,
[tex]\[ 9^{\frac{1}{2}} = 3 \][/tex]
Step 2: Apply the Product Property of Exponents to confirm [tex]\( \left(9^{\frac{1}{2}}\right)^2 \)[/tex].
We know from the exponent properties that:
[tex]\[ (a^{m})^{n} = a^{m \cdot n} \][/tex]
Here [tex]\(a = 9\)[/tex], [tex]\(m = \frac{1}{2}\)[/tex], and [tex]\(n = 2\)[/tex]:
[tex]\[ \left(9^{\frac{1}{2}}\right)^2 = 9^{\frac{1}{2} \cdot 2} = 9^{1} = 9 \][/tex]
So,
[tex]\[ \left(9^{\frac{1}{2}}\right)^2 = 9 \][/tex]
### Part (b)
Another number that can be squared to result in 9 is the negative counterpart of [tex]\( 3 \)[/tex]. This is because squaring a negative number also results in the positive square value.
Step 1: Consider [tex]\(-3\)[/tex]:
When you square [tex]\(-3\)[/tex], you get:
[tex]\[ (-3)^2 = (-3) \cdot (-3) = 9 \][/tex]
Thus, [tex]\(-3\)[/tex] is another number that, when squared, results in 9.
### Summary
From the steps above:
1. The result of [tex]\( 9^{\frac{1}{2}} \)[/tex] is 3.
2. Squaring this result indeed confirms [tex]\(\left(9^{\frac{1}{2}}\right)^2 = 9 \)[/tex].
3. Another number, [tex]\(-3\)[/tex], when squared, also yields 9 since [tex]\((-3)^2 = 9\)[/tex].
So, the detailed step-by-step explorations confirm both [tex]\(3\)[/tex] and [tex]\(-3\)[/tex] as solutions for numbers that when squared give [tex]\(9\)[/tex].
