Section 1.2 : Rational Exponents
For problems 1 – 15 evaluate the given expression and write the answer as a single number with no exponents.
- \({64^{\frac{1}{2}}}\)
- \( - {64^{\frac{1}{2}}}\)
- \({16^{\frac{1}{2}}}\)
- \({16^{\frac{1}{4}}}\)
- \({\left( { - 243} \right)^{\frac{1}{5}}}\)
- \({121^{ - \,\,\frac{1}{2}}}\)
- \({\left( { - 64} \right)^{ - \,\,\frac{1}{3}}}\)
- \({\left( { \displaystyle \frac{{625}}{{256}}} \right)^{\frac{1}{4}}}\)
- \({\left( { \displaystyle - \frac{{27}}{8}} \right)^{\frac{1}{3}}}\)
- \({49^{\frac{5}{2}}}\)
- \({64^{ - \,\,\frac{5}{6}}}\)
- \({\left( { - 729} \right)^{\frac{4}{3}}}\)
- \({\left( { \displaystyle \frac{{121}}{{36}}} \right)^{ - \,\,\frac{3}{2}}}\)
- \({\left( { \displaystyle - \frac{{32}}{{243}}} \right)^{\frac{2}{5}}}\)
- \({\left( { \displaystyle \frac{{81}}{{625}}} \right)^{\frac{3}{4}}}\)
For problems 16 – 23 simplify the given expression and write the answer with only positive exponents.
- \({\left( {{p^{ - 2}}{q^{ - 4}}} \right)^{\frac{3}{2}}}\)
- \({x^{\frac{3}{4}}}{\left( {{x^2}\,{x^{ - \,\,\frac{1}{3}}}} \right)^{\frac{3}{2}}}\)
- \({a^{\frac{1}{2}}}\,{a^{ - \,\,\frac{1}{3}}}\,{a^{\frac{1}{4}}}\)
- \({\left( {{m^{ - \,\,\frac{7}{3}}}{n^{\frac{5}{4}}}} \right)^{ - \,\,\frac{8}{9}}}\)
- \({\left( { \displaystyle \frac{{{a^{ - \,\,\frac{1}{3}}}\,{b^2}}}{{{b^{\frac{2}{3}}}\,{a^{ - \,\,\frac{3}{4}}}}}} \right)^{\frac{1}{5}}}\)
- \({\left( { \displaystyle \frac{{{p^{\frac{1}{2}}}\,{q^{\frac{1}{3}}}}}{{{p^{ - \,\,\frac{1}{3}}}{q^{ - \,\,\frac{1}{4}}}}}} \right)^{ - 3}}\)
- \({\left( { \displaystyle \frac{{{x^{\frac{3}{4}}}\,{y^{ - \,\,\frac{2}{3}}}}}{{{x^{\frac{7}{4}}}}}} \right)^{\frac{7}{8}}}\)
- \({\left( { \displaystyle \frac{{{b^3}\,{c^{ - \,\,\frac{1}{4}}}\,{a^{ - 1}}}}{{{b^{\frac{1}{4}}}\,{a^{ - \,\,\frac{2}{7}}}{c^{\frac{3}{2}}}}}} \right)^{\frac{2}{3}}}\)
For problems 24 & 25 determine if the statement is true or false. If it is false explain why it is false and give a corrected version of the statement.
- \({a^{ - \,\,\frac{3}{2}}} = {a^{\frac{2}{3}}}\)
- \({x^{ - n}} = {x^{\frac{1}{n}}}\)