Let's analyze the problem step-by-step to determine if the statement [tex]\( b^{1 / n} = \sqrt{6} \)[/tex] is true under the given conditions.
Given:
1. [tex]\( b \)[/tex] is a nonnegative real number.
2. [tex]\( n \)[/tex] is a positive integer.
The problem states:
[tex]\[ b^{\frac{1}{n}} = \sqrt{6} \][/tex]
To verify the truth value of this equation, we need to assess if there exist values of [tex]\( b \)[/tex] and [tex]\( n \)[/tex] that satisfy it. Let's explore this step by step:
1. Let [tex]\( \sqrt{6} \)[/tex] be expressed in another form:
[tex]\[ \sqrt{6} = 6^{\frac{1}{2}} \][/tex]
2. In the general form:
[tex]\[ b^{\frac{1}{n}} = 6^{\frac{1}{2}} \][/tex]
For the equation to hold true, [tex]\( b \)[/tex] must be chosen such that raising it to the power of [tex]\( \frac{1}{n} \)[/tex] yields [tex]\( 6^{\frac{1}{2}} \)[/tex].
3. Comparing the exponents on both sides of the equation, we can set:
[tex]\[ b^{\frac{1}{n}} = 6^{\frac{1}{2}} \][/tex]
4. By raising both sides to the power of [tex]\( n \)[/tex], we get:
[tex]\[ b = (6^{\frac{1}{2}})^n \][/tex]
5. Simplifying the right-hand side, we get:
[tex]\[ b = 6^{\frac{n}{2}} \][/tex]
We see that if we let:
[tex]\[ n = 2 \][/tex]
and,
[tex]\[ b = 6 \][/tex]
Then:
[tex]\[ 6^{\frac{1}{2}} = \sqrt{6} \][/tex]
Given these values:
- [tex]\( n = 2 \)[/tex] (a positive integer)
- [tex]\( b = 6 \)[/tex] (a nonnegative real number)
This satisfies the original assertion:
[tex]\[ b^{\frac{1}{n}} = \sqrt{6} \][/tex]
Therefore, the statement [tex]\( b^{\frac{1}{n}} = \sqrt{6} \)[/tex] can be confirmed as true under the appropriate selection of [tex]\( b \)[/tex] and [tex]\( n \)[/tex]:
The correct answer is:
A. True
