How do you simplify $\sqrt[7]{x^{2}}$ $root7(x^2)$ and write it in exponential form?

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Answer 1 · 2018-03-12T11:34:38

$x^{\frac{2}{7}}$ $x^(2/7$

Before we start, let's revise the exponent rules,

Product rule: $a^{x} \times a^{y} = a^{x + y}$ $a^x xxa^y=a^(x+y$
Quotient rule: $a^{x} \div a^{y} = a^{x - y}$ $a^x -:a^y=a^(x-y$
Power rule: ${(a^{x})}^{y} = a^{x y}$ $(a^x)^y=a^(xy)$
Power of a product rule: ${(a b)}^{x} = a^{x} \times b^{x}$ $(ab)^x=a^x xx b^x$
Power of a quotient rule: ${(\frac{a}{b})}^{x} = \frac{a^{x}}{b^{x}}$ $(a/b)^x=(a^x)/(b^x)$
Zero exponent: $a^{0} = 1$ $a^0=1$
Negative exponent: $a^{- x} = \frac{1}{a^{x}}$ $a^-x=1/a^x$
Fractional exponent: $a^{\frac{x}{y}} = \sqrt[y]{a^{x}}$ $a^(x/y)=root(y)(a^x)$

Now let's begin,

$\sqrt[7]{x^{2}}$ $root(7)(x^2)$

Using rule 8 - Fractional exponent, we get,

$x^{\frac{2}{7}}$ $x^(2/7$

How do you simplify 7√x2root7(x^2) and write it in exponential form?