Which is equivalent to RootIndex 5 StartRoot 1,215 EndRoot Superscript x?

243x
1,215 Superscript one-fifth x
1,215 Superscript StartFraction 1 Over 5 x EndFraction
243 Superscript StartFraction 1 Over x EndFraction

Respuesta :

Answer:

1,215 Superscript one-fifth x

Step-by-step explanation:

Given:

The expression to simplify is given as:

[tex](\sqrt[5]{1215})^x[/tex]

We know that,

[tex]\sqrt[n]{a} = a^{\frac{1}{n}}[/tex]

Here, [tex]n=5, a=1215[/tex]

So, [tex]\sqrt[5]{1215} = 1215^{\frac{1}{5}}[/tex]

So, the above expression becomes:

[tex](\sqrt[5]{1215})^x = (1215^{\frac{1}{5}})^x[/tex]

Now, using the law of indices [tex](a^m)^n=a^{(m\times n)}[/tex]

Here, [tex]a=1215,m=\frac{1}{5},n=x[/tex]

So, the expression is finally simplified to;

[tex]=(1215)^{({\frac{1}{5}}\times x)}\\\\=(1215)^{\frac{1}{5}x}[/tex]

Therefore, the second option is the correct one.

[tex](\sqrt[5]{1215})^x[/tex] = 1,215 Superscript one-fifth x

kayym

Answer:

b. 1,215 Superscript one-fifth x

Step-by-step explanation: