Section 1.2 : Rational Exponents
6. Evaluate the following expression and write the answer as a single number without exponents.
\[{\left( {\frac{8}{{343}}} \right)^{ - \frac{2}{3}}}\]Show All Steps Hide All Steps
Hint : Don’t forget your basic exponent rules and how the first two practice problems worked.
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 2Now, 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,
\[{\left( {\frac{8}{{343}}} \right)^{ - \frac{2}{3}}} = {\left( {\frac{{343}}{8}} \right)^{\,\frac{2}{3}}} = \frac{{{{343}^{\frac{2}{3}}}}}{{{8^{\frac{2}{3}}}}} = \frac{{{{\left( {{{343}^{\frac{1}{3}}}} \right)}^2}}}{{{{\left( {{8^{\frac{1}{3}}}} \right)}^2}}} = \frac{{{7^2}}}{{{2^2}}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{49}}{4}}}\]