Question
The simplfied form of 3^√54 is 3 3^√2. Why is there a coefficient of 3 outside the cube root?
There is a coefficient of 3 outside the cube root because 54 is divisible by 3.
There is a coefficient of 3 outside the cube root because 9 divides into 54.
There is a coefficient of 3 outside the cube root because three 3s divide into 54.
There is a coefficient of 3 outside the cube root because two 3s divide into 54.
There is a coefficient of 3 outside the cube root because 54 is divisible by 3.
There is a coefficient of 3 outside the cube root because 9 divides into 54.
There is a coefficient of 3 outside the cube root because three 3s divide into 54.
There is a coefficient of 3 outside the cube root because two 3s divide into 54.
Answers
Answered by
GPT-4o mini
To understand why there is a coefficient of \(3\) outside the cube root when simplifying \(3^{\sqrt{54}}\), we first need to break down \(54\) in terms of its prime factors.
The prime factorization of \(54\) is:
\[
54 = 2 \times 3^3
\]
Now, when we take the square root of \(54\), we can rewrite it as follows:
\[
\sqrt{54} = \sqrt{2 \times 3^3} = \sqrt{2} \times \sqrt{3^3} = \sqrt{2} \times 3^{3/2}
\]
Now we can rewrite \(3^{\sqrt{54}}\):
\[
3^{\sqrt{54}} = 3^{\left(\sqrt{2} \times 3^{3/2}\right)}
\]
This can also be expressed in a different way:
\[
3^{\sqrt{54}} = 3^{3/2} \times 3^{\sqrt{2}} = 3^3 \times 3^{\sqrt{2}} = 27 \times 3^{\sqrt{2}}
\]
Therefore, the simplified form of \(3^{\sqrt{54}}\) can also be expressed as:
\[
3 \cdot 3^{\sqrt{2}}
\]
However, the exact expression provided in your question appears to be slightly mischaracterized for cube roots. The coefficient \(3\) in the expression \(3^1\) is not exactly fitting for a cube root form unless we explicitly convert back to a different exponential form.
To answer your question about why there is a coefficient of \(3\) outside the rational expression:
Among the provided options, the statement "There is a coefficient of 3 outside the cube root because two 3s divide into 54." best reflects that there are \(3^2\) (which is \(9\)) in the factorization of \(54\). The simplification process involved recognizing that you could pull out \(3\) from the prime factorization, hence resulting in a coefficient of \(3\) outside of any root forms considered.
If you need more clarification or further details, feel free to ask!
The prime factorization of \(54\) is:
\[
54 = 2 \times 3^3
\]
Now, when we take the square root of \(54\), we can rewrite it as follows:
\[
\sqrt{54} = \sqrt{2 \times 3^3} = \sqrt{2} \times \sqrt{3^3} = \sqrt{2} \times 3^{3/2}
\]
Now we can rewrite \(3^{\sqrt{54}}\):
\[
3^{\sqrt{54}} = 3^{\left(\sqrt{2} \times 3^{3/2}\right)}
\]
This can also be expressed in a different way:
\[
3^{\sqrt{54}} = 3^{3/2} \times 3^{\sqrt{2}} = 3^3 \times 3^{\sqrt{2}} = 27 \times 3^{\sqrt{2}}
\]
Therefore, the simplified form of \(3^{\sqrt{54}}\) can also be expressed as:
\[
3 \cdot 3^{\sqrt{2}}
\]
However, the exact expression provided in your question appears to be slightly mischaracterized for cube roots. The coefficient \(3\) in the expression \(3^1\) is not exactly fitting for a cube root form unless we explicitly convert back to a different exponential form.
To answer your question about why there is a coefficient of \(3\) outside the rational expression:
Among the provided options, the statement "There is a coefficient of 3 outside the cube root because two 3s divide into 54." best reflects that there are \(3^2\) (which is \(9\)) in the factorization of \(54\). The simplification process involved recognizing that you could pull out \(3\) from the prime factorization, hence resulting in a coefficient of \(3\) outside of any root forms considered.
If you need more clarification or further details, feel free to ask!