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If you are looking for some problems with solutions you can find some by clicking on the "Practice Problems" link above.
If you are looking for some problems with solutions you can find some by clicking on the "Practice Problems" link above.
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.
- \(\displaystyle \frac{1}{{6x}} = 6{x^{ - 1}}\)
- \({\left( {{x^3}} \right)^7} = {x^{10}}\)
- \({\left( {{m^3}{n^4}} \right)^2} = {m^{12}}{n^8}\)
- \({\left( {{{\left( {{z^2}} \right)}^3}} \right)^4} = {z^{24}}\)
- \({\left( {x + y} \right)^3} = {x^3} + {y^3}\)