cadaVR anatomy by Mozaik Education

Exponentiation and computing the roots (reminder)

In our earlier studies, we defined the powers of real numbers with positive integer and later on with integer indices. We familiarised ourselves with the concept of the square root and higher roots. We are brushing up on the identities of these two operations below.

The identities of powers

The identities of roots

\latex{a^{m}\times a^{n}= a^{m+n}, a\in \R ; a\neq 0; m,n \in \Z}

\latex{\frac{a^{m} }{a^{n} }= a^{m-n} , a\in \R; a\neq 0; m,n\in \Z}

\latex{(a\times b)^{n}= a^{n}\times b^{n}, a,b\in \R; a,b\neq 0; n\in \Z}

\latex{(\frac{a}{b} )^{n}= \frac{a^{n} }{b^{n} } , a,b\in \R; a,b\neq 0; n\in \Z}

\latex{(a^{n})^{m}= a^{n\times m} , a\in \R; a,b\neq 0; n\in \Z}

\latex{\sqrt[m]{a\times b} = \sqrt[n]{a} \times \sqrt[n]{b} , a,b\geq 0; n\in \N; n\geq 2}

\latex{\sqrt[n]{\frac{a}{b} } = \frac{\sqrt[n]{a} }{\sqrt[n]{b} } , a\geq 0, b\gt 0; n\in \N; n\geq 2}

\latex{\sqrt[n]{a^{k} } = (\sqrt[n]{a} ^{})^{k}, a\gt 0; n\in \N; n\geq 2; k\in \Z}

\latex{\sqrt[n]{\sqrt[m]{a} } = \sqrt[n\times m]{a} , a\geq 0; m,n\in \N; m,n\geq 2}

\latex{\sqrt[n]{a^{m} } = \sqrt[n\times k]{a^{m\times k} } , a\gt 0; n,k\in \N; n,k\geq 2, m\in \Z}

\latex{ A })

\latex{ B })

\latex{ C })

\latex{ D })

\latex{ E })

\latex{ I. }

\latex{ II. }

\latex{ III. }

\latex{ IV. }

\latex{ V. }

Example 1

Let us simplify the following expressions with the help of thepowers identities:

\latex{\frac{{{{({a^2})}^{ - 3}}\, \times {a^5}\, \times {{({a^{ - 1}})}^{ - 2}}}}{{{{({a^5})}^{ - 4}}\, \times {a^{ - 6}}}},\;\;\;a \ne 0;}
\latex{\frac{{{{({a^{ - 3}}\, \times b)}^{ - 4}}\, \times {{({a^2}\, \times {b^{ - 3}})}^{ - 5}}}}{{{{({b^4})}^{ - 2}}\, \times {{({a^3}\, \times {b^{ - 3}})}^{ - 2}}}},\;\;\;a \ne 0,\;\;b \ne 0;}
\latex{{\left( {\frac{{{x^{ - 3}}}}{{{y^{ - 1}}}}} \right)^{\,5}} \times \;\,\frac{{{{({x^2}\, \times {y^{ - 3}})}^3}}}{{{{({y^2})}^{ - 2}}}}\;\,:\;\,{\left( {\frac{{{y^{ - 2}}\, \times x}}{{{x^{ - 2}}\, \times {y^3}}}} \right)^{ - 3}},\;\;\;x \ne 0,\;\;y \ne 0.}

Solution (a)

\latex{\frac{{{{({a^2})}^{ - 3}}\, \times {a^5}\, \times {{({a^{ - 1}})}^{ - 2}}}}{{{{({a^5})}^{ - 4}}\, \times {a^{ - 6}}}}\;\; = \;\;\frac{{{a^{ - 6}}\, \times {a^5}\, \times {a^2}}}{{{a^{ - 20}}\, \times {a^{ - 6}}}}\;\; = \;\;\frac{{{a^1}}}{{{a^{ - 26}}}}\;\; = \;\;{a^{27}};}

Identity V

Identity I

Identity II

Solution (b)

\latex{\frac{{{{({a^{ - 3}}\, \times b)}^{ - 4}}\, \times {{({a^2}\, \times {b^{ - 3}})}^{ - 5}}}}{{{{({b^4})}^{ - 2}}\, \times {{({a^3}\, \times {b^{ - 3}})}^{ - 2}}}}\;\, = \;\,\frac{{{a^{12}}\, \times {b^{ - 4}}\, \times {a^{ - 10}}\, \times {b^{15}}}}{{{b^{ - 8}}\, \times {a^{ - 6}}\, \times {b^6}}}\;\, = \;\,\frac{{{a^2}\, \times {b^{11}}}}{{{a^{ - 6}}\, \times {b^{ - 2}}}}\;\, = \;\,{a^8}\, \times {b^{13}};}

Identity III and V

Identity I

Identitiy II

Solution (c)

\latex{(\frac{x^{-3} }{y^{-1} } )^{5}\times \frac{(x^{2}\times y^{-3} )^{-3}}{(y^{2} )^{-2}} \div (\frac{y^{-2}\times x }{x^{2}\times y^{3} } )^{-3}= \frac{x^{-15} \times x^{6}\times y^{-9} }{y^{-5} \times y^{-4} } \div \frac{y^{6}\times x^{-3} }{x^{6} \times y^{-9} } = \frac{x^{-9} \times y^{-9} }{y^{-9} } \div \frac{y^{15} }{x^{9} } = x^{-9} \times \frac{x^{9} }{y^{15} } = \frac{1}{y^{15} } = y^{-15}}

Identity III, IV and V

Identitiy I

Identitiy II

Example 2

Let us simplify the following expressions with the help of the roots identities:

\latex{\frac{{\sqrt[5]{{{a^4}}} \times \sqrt[5]{{{a^{ - 1}}}} \times \sqrt[5]{{{a^{ - 2}}}}}}{{\sqrt[5]{{{a^{ - 3}}}} \times \sqrt[5]{{{a^2}}}}},\;\;\;a \ne 0;}

\latex{\frac{{\sqrt[3]{{\sqrt a }} \times \sqrt[5]{{{a^{ - 4}}}} \times \sqrt {{a^5}} }}{{\sqrt[5]{{{a^3}}} \times \sqrt[3]{{{a^{ - 2}}}} \times \sqrt a }},\;\;\;a \gt 0;}

\latex{\frac{{\sqrt[4]{{{a^{ - 2}} \times {b^3}}} \times \sqrt[3]{{{a^2} \times {b^{ - 1}}}}}}{{\sqrt {{a^{ - 1}} \times b} \times \sqrt[6]{{{a^{ - 2}} \times {b^5}}}}},\;\;\;a \gt 0,\;\;b \gt 0.\,}

Solution (a)

\latex{\frac{{\sqrt[5]{{{a^4}}} \times \sqrt[5]{{{a^{ - 1}}}} \times \sqrt[5]{{{a^{ - 2}}}}}}{{\sqrt[5]{{{a^{ - 3}}}} \times \sqrt[5]{{{a^2}}}}}\;\; = \;\;\frac{{\sqrt[5]{{{a^4} \times {a^{ - 1}}{a^{ - 2}}}}}}{{\sqrt[5]{{{a^{ - 3}} \times {a^2}}}}}\;\; = \;\;\sqrt[5]{{\frac{a}{{{a^{ - 1}}}}}}\;\; = \;\;\sqrt[5]{{{a^2}}};}

Identity A)

Identity B)

Solution (b)

\latex{\frac{\sqrt[3]{\sqrt{a} } \times \sqrt[5]{a^{-4} } \times \sqrt{a^{5} } }{\sqrt[5]{a^{3} }\times \sqrt[3]{a^{-2} } \times \sqrt{a} } = \frac{\sqrt[6]{a} \times \sqrt[5]{a^{-4} } \times \sqrt{a^{5} } }{\sqrt[5]{a^{3} }\times \sqrt[3]{a^{-2} }\times \sqrt{a} } \\}

\latex{= \frac{\sqrt[30]{a^{5} } \times \sqrt[30]{(a^{-4})^{6}}\times \sqrt[30]{(a^{5})^{15}} }{\sqrt[30]{(a^{3})^{6}}\times \sqrt[30]{(a^{-2})^{10}}\times \sqrt[30]{a^{15} } }\\}

\latex{= \frac{\sqrt[30]{a^{5}\times a^{-24}\times a^{75} } }{\sqrt[30]{a^{18}\times a^{-20}\times a^{15} } } = \sqrt[30]{\frac{a^{56} }{a^{13} } } = \sqrt[30]{a^{43} } = a\times \sqrt[30]{a^{13} }}

Identity D)

Identity E)

Identity A)

Identity B)

Solution (c)

\latex{\frac{\sqrt[4]{a^{-2}\times b^{2} } \times \sqrt[3]{a^{2}\times b^{-1} } }{\sqrt{a^{-1}\times b }\times \sqrt[6]{a^{2}\times b^{5} } } = \frac{\sqrt[12]{a^{-6} \times b^{9} }\times \sqrt[12]{a^{8}\times b^{-4} } }{\sqrt[12]{a^{-6}\times b^{6} }\times \sqrt[12]{a^{-4}\times b^{10} } } = \frac{\sqrt[12]{a^{2} \times b^{5} } }{\sqrt[12]{a^{-10} \times b^{16} } } \\}

\latex{= \sqrt[12]{a^{12}\times b^{-11} } = a\times \sqrt[12]{b^{-11} }}

Identity E)

Identity B)

Example 3

Which number is greater:

\latex{72^{3}} or \latex{432^{2}} ;

\latex{2\times \sqrt[5]{324}} or \latex{3\times \sqrt[5]{64}}?

Solution (a)

Let us prime factorise the bases:
\latex{72= 2^{3}\times 3^{2}} and \latex{432= 2^{4} \times 3^{3}} .

Expressing the original powers using these:
\latex{72^{3} = (2^{3}\times 3^{2} )^{3}= 2^{9} \times 3^{6}} , and \latex{432^{2} = (2^{4}\times 3^{3} )^{2}= 2^{8} \times 3^{6}} .

The result shows that \latex{72^{3} \gt 432^{2}} .

\latex{ 72 }

\latex{ 36 }

\latex{ 18 }

\latex{ 9 }

\latex{ 3 }

\latex{ 1 }

\latex{ 2 }

\latex{ 3 }

\latex{ 2 }

\latex{ 3 }

\latex{ 432 }

\latex{ 216 }

\latex{ 108 }

\latex{ 54 }

\latex{ 27 }

\latex{ 9 }

\latex{ 3 }

\latex{ 1 }

Solution (b)

Let us prime factorise the numbers under the root sign:
\latex{324= 2^{2} \times 3^{4}} and \latex{64= 2^{6}}
Let us also write the factors from in front of the root signs under the root sign for comparison:
\latex{2\times \sqrt[5]{324} = \sqrt[5]{2^{5}\times 2^{2} \times 3^{4} } = \sqrt[5]{2^{7} \times 3^{4} }} and \latex{3\times \sqrt[5]{64} = \sqrt[5]{3^{5} \times 2^{6} }}

Since
\latex{2^{7} \times 3^{4} = 2\times 2^{6} \times 3^{4} \lt 3\times 2^{6} \times 3^{4} = 3^{5} \times 2^{6}} ,we have \latex{2\times \sqrt[5]{324} \lt 3\times \sqrt[5]{64}}

Example 4

Which number is greater: \latex{\sqrt[4]{2}} or \latex{\sqrt[5]{3}}?

Solution I

Let us form roots with identical indices:
\latex{\sqrt[4]{2} = \sqrt[4\times 5]{2^{5} } = \sqrt[20]{32}} , and \latex{\sqrt[5]{3} = \sqrt[5\times 4]{3^{4} } = \sqrt[20]{81}}

Since \latex{32\lt 81}, we have \latex{\sqrt[4]{2} \lt \sqrt[5]{3}}

Solution II

Let us write the 20th powers of the numbers.
\latex{(\sqrt[4]{2} )^{20}=[ (\sqrt[4]{2} )^{4}]^{5}= 2^{5}} , and \latex{(\sqrt[5]{3} )^{20}= [ (\sqrt[5]{3} ^{5} )]^{4}= 3^{4}}

It is true for the 20th powers of the numbers that \latex{2^{5} \lt 3^{4} } , therefore \latex{\sqrt[4]{2} \lt \sqrt[5]{3}}

Example 5

Let us calculate the values of the following expressions by using the identities:

\latex{\frac{12^{-3}\times 15^{-5} }{10^{-5} \times 18^{-3}\times 9^{-1} }}

\latex{\frac{\sqrt[4]{81}\times \sqrt[3]{15^{5} } }{\sqrt[3]{10}\times \sqrt[3]{18^{2} } \times \sqrt[3]{225^{2} } }}

\latex{\frac{\sqrt[4]{10^{-3} }\times \sqrt[3]{\frac{1}{81} }\times \sqrt[4]{\frac{1}{5^{9} } } }{\sqrt[4]{0,125} \times \sqrt{25^{-2} }\times \sqrt[3]{15^{-4} }\times \sqrt[3]{\frac{1}{25} } }}

Solution (a)

Let us give the bases as powers of prime factors:

\latex{\frac{12^{-3}\times 15^{-5} }{10^{-5}\times 18^{-3} \times 9^{-1} } =\frac{(2^{2}\times 3)^{-3}\times (3\times 5)^{-5}}{(2\times 5)^{-5}\times (2\times 3^{2})^{-3}\times (3^{2})^{-1}} \\}
\latex{= \frac{2^{-6}\times 3^{-3}\times 3^{-5}\times 5^{-5} }{2^{-5} \times 5^{-5}\times 2^{-3} \times 3^{-6} \times 3^{-2} }= \frac{2^{-6}\times 3^{-8}\times 5^{-5} }{2^{-8} \times 3^{-8}\times 5^{-5} } = 2^{2} = 4}

Identity III and V

Identity I

Identity II

Solution (b)

The bases after prime factorisation:

\latex{\frac{\sqrt[4]{81}\times \sqrt[3]{15^{5} } }{\sqrt[3]{10}\times \sqrt[3]{18^{2} } \times \sqrt[3]{225^{2} } } =}\latex{\frac{\sqrt[4]{3^{4} } \times \sqrt[3]{(3\times5 )^{5}} }{\sqrt[3]{2\times 5}\times \sqrt[3]{(2\times 3^{2})^{2}}\times \sqrt[3]{}(3^{2}\times 5 ^{2})^{2} } =}

\latex{= \frac{3\times \sqrt[3]{3^{5}\times 5^{5} } }{\sqrt[3]{2\times 5} \times \sqrt[3]{2^{2}\times 3^{4} } \times \sqrt[3]{3^{4}\times 5^{4} } }=}\latex{ \frac{3\times \sqrt[3]{3^{5} \times 5^{5} } }{\sqrt[3]{2\times 5\times 2^{2} \times 3^{4}\times 3^{4} \times 5^{4} } }=}

\latex{= \frac{\sqrt[3]{3^{3}\times 3^{5}\times ^{5} } }{\sqrt[3]{2^{3}\times 3^{8} \times ^{5} } } = \sqrt[3]{\frac{3^{8}\times 5^{5} }{2^{3}\times 3^{8}\times 5^{5} } }= \sqrt[3]{\frac{1}{2^{3} } }}\latex{= \frac{1}{2}}

Solution (c)

By prime factorising the bases:

\latex{\frac{\sqrt[4]{(2\times 5)^{-3}}\times \sqrt[3]{\frac{1}{3^{4} } } \times \sqrt[4]{5^{-9} } }{\sqrt[4]{\frac{1}{2^{3} } }\times \sqrt{(5^{2})^{-2}} \times \sqrt[3]{(3\times 5)^{-4}}\times \sqrt[3]{\frac{1}{5^{2} } } } =}\latex{\\ \frac{\sqrt[4]{2^{-3}\times 5^{-3} }\times \sqrt[3]{3^{-4} }\times \sqrt[4]{5^{-9} } }{\sqrt[4]{2^{-3} } \times \sqrt{5^{-4} } \times \sqrt[3]{3^{-4} }\times 5^{-4} \times \sqrt[3]{5^{-2} } }=}

\latex{= \frac{\sqrt[4]{2^{-3} }\times \sqrt[4]{5^{-3}\times 5^{-9} } \times \sqrt[3]{3^{-4}} }{\sqrt[4]{2^{-3} }\times 5^{-2} \times \sqrt[3]{5^{-4}\times 5^{-2} } \times \sqrt[3]{3^{-4}} } = \frac{\sqrt[4]{5^{-12} } }{5^{-2} \times \sqrt[3]{5^{-6} } }=} \latex{\frac{5^{-3} }{5^{-2}\times 5^{-2} } = \frac{5^{-3} }{5^{-2}\times 5^{-2} } = \frac{5^{-3} }{5^{-4} } = 5}

Exercises

{{exercise_number}}. Simplify the following expressions with the help of the powers identities:

\latex{\frac{(a^{3})^{2}\times a^{-4} }{(a^{-2})^{3}\times a^{5} }}

\latex{\frac{(b^{-3})^{4}\times (b^{-1})^{-2}}{b^{-3} \times (b^{2})^{2} }}

\latex{\frac{(a^{-5})^{7}\times (a^{3})^{-2}\times (a^{-3})^{-8}}{a^{5}\times (a^{6})^{5}\times a^{-4} }}

\latex{\frac{(a^{-5})^{7}\times (a^{3})^{-2}\times (a^{-3})^{-8}}{a^{5}\times (a^{6})^{5}\times a^{-4} }}

\latex{(\frac{a^{-3}}{b^{-4}})^{-2}\times (\frac{b^{-1}}{a^{4}})^{-3}\div (\frac{a^{-2}}{b^{-3}})^{-5}}

\latex{\frac{(a^{-2}\times b^{-3}\times c^{2} )^{-4}\div (a^{-3}\times b^{-2}\times c^{-2} )^{5} }{(a^{-1}\times b^{-4}\times c^{-3} )^{3}\div (a^{-4}\times b^{-1}\times c^{-3} )^{-5}}}

{{exercise_number}}. Simplify the following expressions with the help of the roots identities:

\latex{\frac{\sqrt{a}\times \sqrt{a^{3} } }{\sqrt{a^{5} } }}

\latex{\frac{\sqrt[3]{b^{2} } \times \sqrt[3]{b^{4} } }{\sqrt[3]{b} }}

\latex{\frac{\sqrt[4]{a^{3} } \times \sqrt[4]{a^{-2} } \times \sqrt[4]{a^{-5} } }{\sqrt[4]{a^{-7} } \times \sqrt[4]{a^{5} } \times \sqrt[4]{a} }}

\latex{\frac{\sqrt[3]{a^{-1} \times b^{-2} } \times \sqrt{a^{3} \times b^{-1} } \times \sqrt[6]{a^{-5} } {} }{\sqrt{a^{-1}\times b^{-3} } \times \sqrt[3]{a^{-3} \times b } \times \sqrt[6]{b} {} }}

\latex{\frac{\sqrt[8]{a^{-5} \times b^{3} } \times \sqrt[3]{a^{-1} \times b^{-2} } \times \sqrt{a^{-1} \times b^{-1} } }{\sqrt{a^{-1} \times b^{-5} } \times \sqrt{a^{3} \times b^{-3} } \times \sqrt[4]{a^{-1}\times b^{3} } } {}}

\latex{\sqrt[5]{\frac{a^{3} }{b^{-4} } } \times \sqrt[6]{\frac{b^{-2} }{a^{-3} } } \times \sqrt[3]{\frac{a^{-1} }{b^{-1} } } \div \sqrt{\frac{b^{-3} }{a^{-5} } } _{}}

{{exercise_number}}. Which number is greater?

\latex{3^{4}} or \latex{\sqrt[4]{3^{17} }}

\latex{10^{-5}} or \latex{32^{-1} \times 625^{-1}}

\latex{\sqrt[3]{120}} or \latex{\sqrt[5]{1800}}

{{exercise_number}}. Calculate the values of the following expressions with the help of the identities of powers and roots:

\latex{\frac{2^{-4} \times 4^{2}\times 8 }{(2^{-1})^{-3}}}

\latex{\frac{3^{5} \times 2^{-3} \times 4^{4} }{6^{4} \times 3}}

\latex{\frac{30^{-4} \times 10^{-5} \times 5^{3} }{60^{-6} \times 2^{2}\times 3 }}

\latex{\frac{147^{-5}\times 75^{-8} \times 105^{-1} \times 7^{-2} }{315^{-4}\times 45^{-3}\times 35^{-10} }}

\latex{\frac{\sqrt[4]{15^{5} }\times \sqrt[4]{3^{3} } \times \sqrt[4]{40^{2} } }{\sqrt[4]{25^{2} }\times \sqrt[4]{20^{3} } }}

\latex{(\sqrt[3]{15} \times \sqrt{20^{-1} } \times \sqrt[12]{2^{-5}\ \times 5^{-7} } )\div (\sqrt[4]{10^{-3} }\times \sqrt[3]{6^{-2} } )}

Puzzle

Calculate the value of the expression \latex{\sqrt{28+ 10\sqrt{3} } + \sqrt{28-10\times \sqrt{3} }} without using a pocket calculator.

Mathematics 11.

-