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Q. 1

Expert-verifiedFound in: Page 282

Book edition
6th

Author(s)
Sullivan

Pages
1200 pages

ISBN
9780321795465

${4}^{3}=\_\_\_\_\_\_\_;{8}^{2/3}=\_\_\_\_\_\_\_;{3}^{-2}=\_\_\_\_\_\_\_$

The answer of ${4}^{3}=64\phantom{\rule{0ex}{0ex}}{8}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=4\phantom{\rule{0ex}{0ex}}{3}^{-2}=\frac{1}{9}$.

We can write ${4}^{3}$ as $4\times 4\times 4$.

The solution is $4\times 4\times 4=64.$

By the laws of exponent, ${a}^{\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$n$}\right.}=\sqrt[n]{{a}^{m}}$.

So,

${8}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=\sqrt[3]{{8}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt[3]{64}\phantom{\rule{0ex}{0ex}}=\sqrt[3]{{\left(4\right)}^{3}}\phantom{\rule{0ex}{0ex}}=4$

By the laws of exponent, ${a}^{-x}=\frac{1}{{a}^{x}}$.

So,

${3}^{-2}=\frac{1}{{3}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{1}{9}$

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