# How do you simplify (4^-1 * 7) ^ 2?

Jul 6, 2018

$\frac{49}{16}$

#### Explanation:

$\text{using the "color(blue)"law of exponents}$

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${\left({4}^{-} 1 \times 7\right)}^{2}$

$= {\left(\frac{1}{4} \times 7\right)}^{2}$

$= {\left(\frac{7}{4}\right)}^{2} = \frac{7}{4} \times \frac{7}{4} = \frac{7 \times 7}{4 \times 4} = \frac{49}{16}$

Jul 6, 2018

$\frac{49}{16}$

#### Explanation:

When we distribute an exponent to another exponent, we multiply the powers. Doing this, we now have

${4}^{- 2} \cdot {7}^{2}$

When we have a negative exponent, we can make it positive by taking it to the denominator. We now have

$\frac{1}{{4}^{2}} \cdot {7}^{2} = \frac{{7}^{2}}{{4}^{2}} = \frac{49}{16}$

Hope this helps!