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Section 1.2 : Rational Exponents

6. Evaluate the following expression and write the answer as a single number without exponents.

\[{\left( {\frac{8}{{343}}} \right)^{ - \frac{2}{3}}}\]

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Hint : Don’t forget your basic exponent rules and how the first two practice problems worked.
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Let’s first recall our basic exponent rules and note that we can easily write this as,

\[{\left( {\frac{8}{{343}}} \right)^{ - \frac{2}{3}}} = {\left( {\frac{{343}}{8}} \right)^{\,\frac{2}{3}}} = \frac{{{{343}^{\frac{2}{3}}}}}{{{8^{\frac{2}{3}}}}} = \frac{{{{\left( {{{343}^{\frac{1}{3}}}} \right)}^2}}}{{{{\left( {{8^{\frac{1}{3}}}} \right)}^2}}}\] Show Step 2

Now, recalling how the first two practice problems worked we can see that,

\[{343^{\frac{1}{3}}} = 7\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}{8^{\frac{1}{3}}} = 2\]

Therefore,

\[{\left( {\frac{8}{{343}}} \right)^{ - \frac{2}{3}}} = {\left( {\frac{{343}}{8}} \right)^{\,\frac{2}{3}}} = \frac{{{{343}^{\frac{2}{3}}}}}{{{8^{\frac{2}{3}}}}} = \frac{{{{\left( {{{343}^{\frac{1}{3}}}} \right)}^2}}}{{{{\left( {{8^{\frac{1}{3}}}} \right)}^2}}} = \frac{{{7^2}}}{{{2^2}}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{49}}{4}}}\]