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Exponentiation to integer index

We want to extend the concept of exponentiation to all integer indices so that the identities proved for the positive integer indices (which make it easier to calculate with powers) would be fulfilled unchanged.

The need for extending the concept is called the principle of constancy.

For example identity II proved in the previous chapter makes it possible
to extend the concept of exponentiation.

The formal application
of identity II
when \latex{ m } < \latex{ n }:

By simplification:

\latex{\frac{7^{3}}{7^{5} } =7^{-2}}

\latex{\frac{7^{3}}{7^{5} } =\frac{1^{}}{7^{2} }}

so let \latex{7^{-2}={\frac{1}{7^{2} }}}.

\latex{\begin{rcases}\\\\\end{rcases}}

DEFINITION: If \latex{n} is a positive integer and \latex{ a\neq 0}, then:

\latex{a^{-n}=\frac{1}{a^{n} }}.

For example:

\latex{2^{-4}=\frac{1}{2^{4} }=\frac{1}{16}};

\latex{5^{-3} = \frac{1}{5^{3} } = \frac{1}{125}} ;

\latex{3^{-5} = \frac{1}{3^{5} } = \frac{1}{243}};

\latex{10^{-4} = \frac{1}{10^{4} } = \frac{1}{10000}};

\latex{\Biggl(-\frac{1}{2}\Biggr) ^{-2}=\frac{1}{\Biggl(-{\dfrac{1}{2}}\Biggr) ^{2}} =\frac{1}{\frac{1}{4} } =4.}

⯁ ⯁ ⯁

A non-zero real number raised to the power of (\latex{ –1 }) is actually equal to
its reciprocal: \latex{a^{-1}= \frac{1}{a}}, if \latex{a\neq 0}

The formal application
of identity II
when \latex{ m } = \latex{ n }:

By simplification:

\latex{\frac{a^{5}}{a^{5} } =a^{0}}

\latex{\frac{a^{5}}{a^{5} } =1}

so let \latex{ a^{0}=1 }.

\latex{\begin{rcases}\\\\\\\\\end{rcases}}

DEFINITION: If \latex{a\neq 0}, then \latex{ a^{0}=1 }.

For example: \latex{ 4^{0}=1 }; \latex{ 1.3^{0}=1 }; \latex{ (-5)^{0}=1 } or \latex{ 10^{0}=1 }.

Note: We would have two options to interpret \latex{ 0^0 }:

Since when \latex{a\neq 0,a=1} , \latex{ a^{0}=1 } , then \latex{ 0^{0} } could be \latex{ 1 };
Since \latex{ 0^{n}=0 } for any positive integer \latex{ n }, then \latex{ 0^{0} } could also be \latex{ 0 }.

For the sake of clarity, \latex{ 0^{0} } is not interpreted.

We introduced exponentiation to negative integer indices so that
identity II stays valid.

It can also be proved that the other exponential identities are valid for
general integer indices using this definition.

\latex{a^{-4}\times a^{-6}=\frac{1}{a^{4}}\times \frac{1}{a^{6} } =\frac{1}{a^{10} } =a^{-10}}

(I)

\latex{\left(a\times b\right) ^{-5} =\frac{1}{{\left(a\times b\right)^{5}}} =\frac{1}{a^{5}\times b^{5} } =\frac{1}{a^{5} } \times \frac{1}{b^{5} } =a^{-5} \times b^{-5}}

(III)

\latex{\left(\frac{a}{b} \right) ^{-5}=\frac{1}{\left(\frac{a}{b} \right)^{5} } =\frac{1}{\frac{a^{5} }{b^{5} } }=\frac{b^{5} }{a^{5} }=\frac{1}{a^{5} }\div \frac{1}{b^{5} } =a^{-5}\div b^{-5}= \frac{a^{-5} }{b^{-5}}}

(IV)

\latex{\left(a^{3} \right) ^{-5}=\frac{1}{\left(a^{3} \right) ^{5}}=\frac{1}{a^{15} }=a^{-15}}

(V)

Example 1

Which number is greater: \latex{A=\frac{15^{-6} }{3^{2} }} or \latex{B=\frac{9^{-4}\times 25^{-3} }{5}?}

Solution

\latex{A=\frac{15^{-6}\ }{3^{2} }=\frac{(3\times 5)^{-6} }{3^{2} } =\frac{3^{-6}\times 5^{-6} }{3^{2} } =5^{-6} \times \frac{3^{-6} }{3^{2} } =5^{-6}\times 3^{-8}}.

\latex{B=\frac{9^{-4}\times 25^{-3} }{5}=\frac{(3^{2}) ^{-4} \times \left(5^{2} \right)^{-3} }{5}= 3^{-8}\times \frac{5^{-6} }{5}=3^{-8}\times 5^{-7}}.

Identity II

Identity III

Identity V

Identity II

Since \latex{5^{-6}= \frac{1}{5^{6} } \gt \frac{1}{5^{7} } =5^{-7}}, then \latex{A\gt B}.

Example 2

Let us calculate the value of \latex{\Biggl(\frac{2^{-3} }{3^{2} }\Biggr) ^{-5} \times \frac{(3^{-1})^{5} }{(2^{2}\times 3^{-3})^{-3}}}.

Solution

\latex{\Biggl(\frac{2^{-3} }{3^{2} }\Biggr) ^{-5} \times \frac{(3^{-1})^{5} }{(2^{2}\times 3^{-3})^{-3} }=\frac{2^{15} }{3^{-10} }\times \frac{3^{-5} }{2^{-6}\times 3^{9} }= \frac{2^{15}\times 3^{-5} }{2^{-6} \times 3^{-1}} =2^{21} \times 3^{-4} =\frac{2^{21} }{3^{4}}}.

Exercises

{{exercise_number}}. Calculate the values of the below expressions.

\latex{ 2^{-3} }

\latex{\left(-3\right) ^{-2} }

\latex{\Biggl(\frac{1}{3}\Biggr) ^{-2} }

\latex{\Biggl(-\frac{2}{3}\Biggr) ^{-1} }

\latex{\left(5^{-3} \right) ^{-2} \times \left(5^{4} \right)^{-3}\times 5^{7} }

\latex{\frac{\left(5^{-2} \right) ^{-3} \times \left(5^{3} \right)^{-5} }{\left(5^{2} \right) ^{3} \times \left(5^{-7} \right)^{2}\ } }

\latex{\left(3^{-1}\times 7^{2} \right) \times \left(7^{-4}\times 3 \right) ^{-3} }

\latex{\left(2^{-2}\times 5 \right)^{-4} \times \left(5^{-2}\times 2^{3} \right) ^{-3} }

\latex{\frac{3^{-5}\times 5^{-3} }{5^{2}\times 3^{-6} } \times \frac{\left(3^{-1} \right) ^{3} }{3^{-3}\times 5^{6} } }

{{exercise_number}}. Do the exponentiations and simplify the expressions for all possible values of the variables.

\latex{\Biggl(-\frac{a}{b}\Biggr)^{-2} }

\latex{\left(2x\right)^{-3} }

\latex{\left(a^{2}b^{-3} \right)^{-2} \times b^{-5} }

\latex{\frac{(a^{-1} )^{3} \times \left(a^{2} \right)^{-3}\times a^{2} }{a^{3}\times (a^{-2} )^{-3} } }

\latex{(2ab^{2})^{-2}\times (b\times a^{-3})^{-4}}

\latex{\frac{(xy^{-3} )^{-2} }{x^{-4}y^{-4} } \times \frac{\left(x^{-1}\right)^{-5}} {\Biggl(\dfrac{1}{y}\Biggr)^{-2} }}

\latex{\frac{\Biggl(\dfrac{a^{-1} }{b^{2} }\Biggr)^{-3}\times \Biggl(\dfrac{b^{-2} }{a}\Biggr)^{3} }{\Biggl(\dfrac{b^{-3} }{a^{2} }\Biggr)^{-2}\times \Biggl(\dfrac{a^{-4}}{b^{7} }\Biggr)^{2}} }

\latex{{\frac{\Biggl(\dfrac{2x^{-5} }{y^{2} }\Biggr)^{-1}\times \Biggl(\dfrac{y^{-3} }{2x^{6} }\Biggr)^{-2} }{\Biggl(\dfrac{2y^{2} }{x^{-3} }\Biggr)^{-3}\times \Biggl(\dfrac{x^{-2}}{2y^{-4} }\Biggr)^{3}} }}

{{exercise_number}}. Express the fractions as a product of powers.

\latex{\frac{27}{10000} }

b) \latex{\frac{512}{81} }

c) \latex{\frac{625}{256} }

{{exercise_number}}. Do the following operations, simplifying as far as possible.

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

\latex{\frac{10^{-3} }{100 } \times 1,000^{2}}

\latex{9^{-4}\times \frac{81^{-1} }{27^{-4} } }

\latex{(49\times 16)^{-3}\times (7^{-4}\times 8^{-2})^{-2}}

\latex{\frac{(32^{-2})^{3}}{1024^{-4}} \times \frac{64^{-3}}{16^{-5}} }

{{exercise_number}}. Which number is greater:

\latex{4^{-3}} or \latex{3^{-4}};

\latex{10^{-7} } or \latex{2^{-6} \times 5^{-8} };

c) \latex{32^{-5}} or \latex{3^{-7} \times (3\times 2^{-4})^{6}};

\latex{3^{7} \times6^{-8} } or \latex{\Biggl(\frac{2}{3}\Biggr) ^{-5}\times 18^{-3}?}

Puzzle

Give numbers \latex{a}, \latex{b}, \latex{c}, \latex{d} so that \latex{a\times 10^{-2}+ b\times 10^{-4}+ c\times 10^{-1}+ d\times 10^{-3}=0.2305}.

Mathematics 9.

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