Section 1.1 : Integer Exponents
For problems 1 – 10 evaluate the given expression and write the answer as a single number with no exponents.
- \(2 \cdot {5^2} + {\left( { - 4} \right)^2}\)
- \({6^0} - {3^5}\)
- \(3 \cdot {4^3} + 2 \cdot {3^2}\)
- \({\left( { - 1} \right)^4} + 2{\left( { - 3} \right)^4}\)
- \({7^0}{\left( {{4^2} \cdot {3^2}} \right)^2}\)
- \( - {4^3} + {\left( { - 4} \right)^3}\)
- \(8 \cdot {2^{ - 3}} + {16^0}\)
- \({\left( {{2^{ - 1}} + {3^{ - 1}}} \right)^{ - 1}}\)
- \(\displaystyle \frac{{{3^2} \cdot {{\left( { - 2} \right)}^3}}}{{{6^{ - 2}}}}\)
- \(\displaystyle \frac{{{4^{ - 2}} \cdot {5^3}}}{{{3^{ - 4}}}}\)
For problems 11 – 18 simplify the given expression and write the answer with only positive exponents.
- \({\left( {3{x^{ - 2}}{y^{ - 4}}} \right)^{ - 1}}\)
- \({\left( {{{\left( {2{a^2}} \right)}^{ - 3}}{b^4}} \right)^{ - 3}}\)
- \(\displaystyle \frac{{{c^{ - 6}}{b^{10}}}}{{{b^9}{c^{ - 11}}}}\)
- \(\displaystyle \frac{{4{a^3}{{\left( {{b^2}a} \right)}^{ - 4}}}}{{{c^{ - 6}}{a^2}{b^{ - 7}}}}\)
- \(\displaystyle \frac{{{{\left( {6{v^2}} \right)}^{ - 1}}{w^{ - 4}}}}{{{{\left( {2v} \right)}^{ - 3}}{w^{10}}}}\)
- \({\left( {\displaystyle \frac{{{{\left( {8{x^{21}}} \right)}^0}{y^{ - 3}}{x^8}}}{{{y^{ - 9}}{x^{ - 1}}}}} \right)^6}\)
- \({\left( {\displaystyle \frac{{{a^2}{b^{ - 4}}{c^{ - 1}}}}{{{b^{ - 9}}{c^8}{a^{ - 4}}}}} \right)^{ - 2}}\)
- \({\left( {\displaystyle \frac{{{p^{ - 6}}{q^7}{{\left( {{p^2}q} \right)}^{ - 3}}}}{{{{\left( {{p^{ - 1}}{q^{ - 4}}} \right)}^2}{p^{10}}}}} \right)^3}\)
For problems 19 – 23 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.
- \(\displaystyle \frac{1}{{6x}} = 6{x^{ - 1}}\)
- \({\left( {{x^3}} \right)^7} = {x^{10}}\)
- \({\left( {{m^3}{n^4}} \right)^2} = {m^{12}}{n^8}\)
- \({\left( {{{\left( {{z^2}} \right)}^3}} \right)^4} = {z^{24}}\)
- \({\left( {x + y} \right)^3} = {x^3} + {y^3}\)