Question

What index should the root form of the expression
$12^{\frac {6}{3}}\cdot 12^{-\frac {1}{2}}$
have?
$\square $

Answer 1

To find the index of the root form of the expression \( 12^{\frac{6}{3}} \cdot 12^{-\frac{1}{2}} \), we first simplify the expression.

1. **Simplify the Exponents**:
\[
\frac{6}{3} = 2
\]
This means:
\[
12^{\frac{6}{3}} = 12^2
\]
Therefore, we rewrite the expression as:
\[
12^2 \cdot 12^{-\frac{1}{2}}
\]

2. **Combine the Exponents**:
We can add the exponents because of the property \( a^m \cdot a^n = a^{m+n} \):
\[
12^2 \cdot 12^{-\frac{1}{2}} = 12^{2 - \frac{1}{2}}
\]

To combine \( 2 \) and \( -\frac{1}{2} \), we convert \( 2 \) to a fraction:
\[
2 = \frac{4}{2}
\]
Now we can write:
\[
2 - \frac{1}{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2}
\]
Thus,
\[
12^{2 - \frac{1}{2}} = 12^{\frac{3}{2}}
\]

3. **Express in Root Form**:
The expression \( 12^{\frac{3}{2}} \) can be expressed in root form as follows:
\[
12^{\frac{3}{2}} = \left(12^3\right)^{\frac{1}{2}} = \sqrt{12^3}
\]
This indicates that the index should be \( 2 \) because we are taking the square root.

Thus, the index of the root form of the expression \( 12^{\frac{6}{3}} \cdot 12^{-\frac{1}{2}} \) is:
\[
\boxed{2}
\]