How do you simplify ${(4^{- 1} \cdot 7)}^{2}$ $(4^-1 * 7) ^ 2$ ?

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How do you simplify ${(4^{- 1} \cdot 7)}^{2}$ $(4^-1 * 7) ^ 2$ ?

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Answer 1 · 2018-07-06T17:26:37

$\frac{49}{16}$ $49/16$

Explanation:

$using the law of exponents$ $"using the "color(blue)"law of exponents"$

$∙ x a^{- m} \Leftrightarrow \frac{1}{a^{m}}$ $•color(white)(x)a^-mhArr1/a^m$

${(4^{- 1} \times 7)}^{2}$ $(4^-1xx7)^2$

$= {(\frac{1}{4} \times 7)}^{2}$ $=(1/4xx7)^2$

$= {(\frac{7}{4})}^{2} = \frac{7}{4} \times \frac{7}{4} = \frac{7 \times 7}{4 \times 4} = \frac{49}{16}$ $=(7/4)^2=7/4xx7/4=(7xx7)/(4xx4)=49/16$

Answer 2 · 2018-07-06T17:41:09

$\frac{49}{16}$ $49/16$

Explanation:

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

$4^{- 2} \cdot 7^{2}$ $4^(-2)*7^2$

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

$\frac{1}{4^{2}} \cdot 7^{2} = \frac{7^{2}}{4^{2}} = \frac{49}{16}$ $1/(4^2)*7^2=(7^2)/(4^2)=49/16$

Hope this helps!

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How do you simplify ${(4^{- 1} \cdot 7)}^{2}$ $(4^-1 * 7) ^ 2$ ?

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