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Power, root, logarithm
Exponentiation to an integer exponent
DEFINITIONS:
\latex{a^{m}=a\times a\times a...\times a} is product with \latex{ m } factors
where \latex{a\in \R;m\in \N^{+}.}
For \latex{m=1}: \latex{a^{1}=a.}
where \latex{a\in \R;m\in \N^{+}.}
For \latex{m=1}: \latex{a^{1}=a.}
\latex{a^{-n}=\frac{1}{a^{n} }; a\neq 0; a\in \R; n\in \natnums^{+}.}
\latex{a^{0}=1; a\neq 0; a\in \R.}
\latex{a^{0}=1; a\neq 0; a\in \R.}
IDENTITIES:
\latex{a^{m}\times a^{n}=a^{m+n};a\in \R; m,n\in \Z.}
\latex{\frac{a^{m} }{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\neq 0;b\neq 0;n\in \Z.}
\latex{\left(\frac{a}{b} \right)^{n}=\frac{a^{n} }{b^{n} };a,b\in \R;a\neq 0;b\neq 0;n\in \Z.}
\latex{(a^{n} )^{m}=a^{n\times m} ;a\in \R;a\neq 0;n,m\in \Z.}
\latex{\left(\frac{a}{b} \right)^{n}=\frac{a^{n} }{b^{n} };a,b\in \R;a\neq 0;b\neq 0;n\in \Z.}
\latex{(a^{n} )^{m}=a^{n\times m} ;a\in \R;a\neq 0;n,m\in \Z.}
Example 1
Which power of \latex{ x } is
- \latex{\frac{(x^{2} )^{4}\times (x^{3} )^{3} }{(x^{4} )^{3}\times x } (x\neq 0);}
- \latex{\frac{(x^{-3} )^{4}\times (x^{-4} )^{-5} }{(x^{3} )^{2}\times (x^{4} )^{-2} } (x\neq 0)?}
Solution (a)
Expand the parentheses and use the identities:
\latex{\frac{(x^{2} )^{4}\times (x^{3} )^{3} }{(x^{4} )^{3}\times x }=\frac{x^{8}\times x^{9} }{x^{12}\times x }=\frac{x^{17} }{x^{13} }=x^{4}.}
Therefore the expression is the fourth power of \latex{ x }.
Solution (b)
Use the previous method:
\latex{\frac{(x^{-3} )^{4}\times (x^{-4} )^{-5} }{(x^{3} )^{2}\times (x^{4} )^{-2} }=\frac{x^{-12}\times x^{20} }{x^{6}\times x^{-8} }=\frac{x^{8} }{x^{-2} }=x^{10}.}
Therefore the expression is the tenth power of \latex{ x }.
Example 2
Find the simplified form of the fraction \latex{\frac{20^{4}\times 21^{3}\times 75 }{2\times 10^{4}\times 14^{3}\times 15^{3} }.}
Solution
Express the bases as the product of prime numbers and use the identities of exponentiation:
\latex{\frac{20^{4}\times 21^{3}\times 75 }{2\times 10^{4}\times 14^{3}\times 15^{3} }=\frac{(2^{2}\times 5 )^{4}\times (3\times 7)^{3}\times (3\times 5^{2} ) }{2\times (2\times 5)^{4}\times (2\times 7)^{3}\times (3\times 5)^{3} }=}
\latex{=\frac{2^{8}\times 5^{4}\times 3^{3}\times 7^{3}\times 3\times 5^{2} }{2\times 2^{4}\times 5^{4}\times 2^{3}\times 7^{3}\times 3^{3}\times 5^{3} }=\frac{2^{8}\times 3^{4}\times 5^{6}\times 7^{3} }{2^{8}\times 3^{3}\times 5^{7}\times 7^{3} }=\frac{3}{5}.}
\latex{=\frac{2^{8}\times 5^{4}\times 3^{3}\times 7^{3}\times 3\times 5^{2} }{2\times 2^{4}\times 5^{4}\times 2^{3}\times 7^{3}\times 3^{3}\times 5^{3} }=\frac{2^{8}\times 3^{4}\times 5^{6}\times 7^{3} }{2^{8}\times 3^{3}\times 5^{7}\times 7^{3} }=\frac{3}{5}.}
\latex{ 20\\ 10\\ 5\\ 1 }
\latex{ 2\\ 2\\ 5 }
\latex{ 2\\1\\ 7\\ 1 }
\latex{ 3\\ 7 }
\latex{ 75\\ 25\\ 5\\ 1 }
\latex{ 3\\ 5\\ 5 }
Example 3
Compute the gravitational force between the Earth and the Moon if the law
of force is: \latex{F=\gamma \times \frac{m_{1}\times m_{2} }{r^{2} },} where \latex{\gamma =6.67\times 10^{-11}\frac{N\times m^{2} }{kg^{2} },} \latex{m_{1}} and \latex{m_{2}} are the masses of the two bodies and \latex{ r } is their distance. (Mass of the Earth: \latex{m_{E}=5.974\times 10^{24}\,kg}, mass of the Moon: \latex{m_{M}=7.347\times 10^{22}\,kg}, the average distance between them is \latex{ 384,404\, km }.)
Solution
Express the distance in meters and substitute all data into the formula using normalized notation.
\latex{F=6.67\times 10^{-11}\frac{N\times m^{2} }{kg^{2} }\times \frac{5.947\times 10^{24}kg\times 7.347\times 10^{22}kg }{(3.844\times 10^{8}m )^{2} }=}
\latex{=\frac{6.67\times 5.974\times 7.347\times 10^{35} }{3.844^{2}\times 10^{16} }N=19.81\times 10^{19}N=1.981\times 10^{20}N.}
\latex{=\frac{6.67\times 5.974\times 7.347\times 10^{35} }{3.844^{2}\times 10^{16} }N=19.81\times 10^{19}N=1.981\times 10^{20}N.}
The \latex{n^{th}} root and exponentiation to a fractional power
DEFINITION: The \latex{2k^{th}(k\in \N^{+} )} root of a non-negative number a is the non-negative number whose \latex{2k^{th}} power equals a.
\latex{(\sqrt[2k]{a})^{2k}=a,} if \latex{a\gt 0.}
\latex{(\sqrt[2k]{a})^{2k}=a,} if \latex{a\gt 0.}
DEFINITION: The \latex{(2k+1)^{th}(k\in \N^{+} )} root of a real number a is the real number whose \latex{(2k+1)^{th}} power equals a.
\latex{(\sqrt[2k+1]{a} )^{2k+1}=a.}
\latex{(\sqrt[2k+1]{a} )^{2k+1}=a.}
IDENTITIES:
\latex{\sqrt[n]{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;k\in \Z;n\in \N;n\geq 2.}
\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;m\in \Z;n,k\in \N;n,k\geq 2.}
\latex{\sqrt[n]{a^{m} }=\sqrt[n\times k]{a^{m\times k} };a\gt 0;m\in \Z;n,k\in \N;n,k\geq 2.}
DEFINITION: The \latex{\frac{m}{n}} th power of the positive number \latex{ a } is the \latex{n^{th}} root of the \latex{m^{th}} power of the base \latex{ a }:
\latex{a^{\frac{m}{n} }=\sqrt[n]{a^{m} },} where \latex{a\gt 0;m\in \Z;n\in \N} and \latex{n\geq 2.}
\latex{a^{\frac{m}{n} }=\sqrt[n]{a^{m} },} where \latex{a\gt 0;m\in \Z;n\in \N} and \latex{n\geq 2.}
Example 4
Compute the following expressions:
- \latex{(\sqrt{3}-\sqrt{2} )\times (3\times \sqrt{2}-\sqrt{3} );}
- \latex{(\sqrt{7}-\sqrt{3} )^{2};}
- \latex{(4\times \sqrt{5}-2\times \sqrt{6} )\times (4\times \sqrt{5}+2\times \sqrt{6} );}
- \latex{(\sqrt{8-2\times \sqrt{7} }+\sqrt{8+2\times \sqrt{7} } )^{2}.}
Solution (a)
Expand the parentheses and use the definition of the square root:
\latex{(\sqrt{3}-\sqrt{2} )\times (3\times \sqrt{2}-\sqrt{3} )=3\times \sqrt{6}-3-3\times 2+\sqrt{6}=}
\latex{=4\times \sqrt{6}-9.}
\latex{=4\times \sqrt{6}-9.}
Solution (b)
Apply the identity for the square of the difference:
\latex{(\sqrt{7}-\sqrt{3} )^{2}=(\sqrt{7} )^{2}-2\times \sqrt{7}\times \sqrt{3}+(\sqrt{3} )^{2}=}
\latex{=7-2\times \sqrt{21}+3=10-2\times \sqrt{21}.}
\latex{=7-2\times \sqrt{21}+3=10-2\times \sqrt{21}.}
Solution (c)
Apply the law for the product of the sum and the difference:
\latex{(4\times \sqrt{5}-2\times \sqrt{6} )\times (4\times \sqrt{5}+2\times \sqrt{6} )=(4\times \sqrt{5} )^{2}-(2\times \sqrt{6} )^{2}=}
\latex{16\times 5-4\times 6=80-24=56.}
\latex{16\times 5-4\times 6=80-24=56.}
Solution (d)
After taking the square of the sum, move the twofold product under one radical sign:
\latex{(\sqrt{8-2\times \sqrt{7} }+\sqrt{8+2\times \sqrt{7} } )^{2}=}
\latex{=8-2\times \sqrt{7}+2\times \sqrt{8-2\times \sqrt{7} }\times \sqrt{8+2\times \sqrt{7} }+8+2\times \sqrt{7}=}
\latex{=16+2\times \sqrt{(8-2\times \sqrt{7} )\times (8+2\times \sqrt{7} )}=16+2\times \sqrt{8^{2}-(2\times \sqrt{7} )^{2} }=}
\latex{=16+2\times \sqrt{64-28}=16+2\times \sqrt{36}=16+12=28.}
\latex{=8-2\times \sqrt{7}+2\times \sqrt{8-2\times \sqrt{7} }\times \sqrt{8+2\times \sqrt{7} }+8+2\times \sqrt{7}=}
\latex{=16+2\times \sqrt{(8-2\times \sqrt{7} )\times (8+2\times \sqrt{7} )}=16+2\times \sqrt{8^{2}-(2\times \sqrt{7} )^{2} }=}
\latex{=16+2\times \sqrt{64-28}=16+2\times \sqrt{36}=16+12=28.}
Example 5
Rationalize the denominator of the following fractions:
- \latex{\frac{3}{\sqrt{6} };}
- \latex{\frac{7}{2-\sqrt{3} };}
- \latex{\frac{10}{\sqrt[4]{5} };}
- \latex{\frac{12}{\sqrt[3]{7}-\sqrt[3]{3} }.}
Solution (a)
Expand the fraction by \latex{\sqrt{6}:}
\latex{\frac{3}{\sqrt{6} }=\frac{3}{\sqrt{6} }\times \frac{\sqrt{6} }{\sqrt{6} }=\frac{3\times \sqrt{6} }{6}=\frac{\sqrt{6} }{2}.}
Solution (b)
Expand the fraction by \latex{2+\sqrt{3}:}
\latex{\frac{7}{2-\sqrt{3} }=\frac{7}{2-\sqrt{3} }\times \frac{2+\sqrt{3} }{2+\sqrt{3} }=\frac{7\times (2+\sqrt{3} )}{2^{2}-(\sqrt{3} )^{2} }=}
\latex{=\frac{7\times (2+\sqrt{3} )}{4-3}=7\times (2+\sqrt{3} ).}
\latex{=\frac{7\times (2+\sqrt{3} )}{4-3}=7\times (2+\sqrt{3} ).}
Solution (c)
Expand the fraction by \latex{\sqrt[4]{5^{3} }:}
\latex{\frac{10}{\sqrt[4]{5} }=\frac{10}{\sqrt[4]{5} }\times \frac{\sqrt[4]{5^{3} } }{\sqrt[4]{5^{3} } }=\frac{10\times \sqrt[4]{5^{3} } }{\sqrt[4]{5^{4} } }=\frac{10\times \sqrt[4]{125} }{5}=2\times \sqrt[4]{125}.}
Solution (d)
Rationalization is done by using the identity \latex{(a-b)\times (a^{2}+ab+b^{2} )=a^{3}-b^{3},} by expanding the fraction with \latex{\sqrt[3]{7^{2} }+\sqrt[3]{21}+\sqrt[3]{3^{2} }:}
\latex{\frac{12}{\sqrt[3]{7}-\sqrt[3]{3} }=\frac{12}{\sqrt[3]{7}-\sqrt[3]{3} }\times \frac{\sqrt[3]{7^{2} }+\sqrt[3]{21}+\sqrt[3]{3^{3} } }{\sqrt[3]{7^{2} }+\sqrt[3]{21}+\sqrt[3]{3^{2} } }=}
\latex{=\frac{12\times (\sqrt[3]{7^{2}+\sqrt[3]{21} }+\sqrt[3]{3^{2} } )}{7-3}=3\times (\sqrt[3]{49}+\sqrt[3]{21}+\sqrt[3]{9} ).}
Example 6
Simplify the following fractions:
- \latex{\frac{x+5\times \sqrt{x} }{7\times \sqrt{x}+35 };}
- \latex{\frac{x+\sqrt{x}-6 }{x-9}.}
Solution (a)
The expression with the square root is valid if \latex{x\geq 0.}
The fraction is valid for every non-negative \latex{ x } as \latex{\sqrt{x}} is also non-negative.
Factorize both the numerator and denominator using that if \latex{x\geq 0,} then \latex{x(\sqrt{x} )^{2}:}
The fraction is valid for every non-negative \latex{ x } as \latex{\sqrt{x}} is also non-negative.
Factorize both the numerator and denominator using that if \latex{x\geq 0,} then \latex{x(\sqrt{x} )^{2}:}
\latex{\frac{x+5\times \sqrt{x} }{7\times \sqrt{x}+35 }=\frac{\sqrt{x}\times (\sqrt{x}+5 ) }{7\times (\sqrt{x}+5 )}=\frac{\sqrt{x} }{7}.}
Solution (b)
The expression is valid if \latex{x\geq 0} and \latex{x\neq 9.}
The numerator is a quadratic expression of \latex{\sqrt{x,}} factorize it using the factored form of quadratic equations.
The roots of the quadratic equation \latex{y^{2}+y-6=0} are \latex{y_{1}=2} and \latex{y_{2}=-3}. The factored form of the left hand side is:
The numerator is a quadratic expression of \latex{\sqrt{x,}} factorize it using the factored form of quadratic equations.
The roots of the quadratic equation \latex{y^{2}+y-6=0} are \latex{y_{1}=2} and \latex{y_{2}=-3}. The factored form of the left hand side is:
\latex{(y-2)\times (y+3)=0.}
In the denominator, we can apply the identity for the difference of two squares to factorize:
\latex{\frac{x+\sqrt{x}-6 }{x-9}=\frac{(\sqrt{x}-2 )\times (\sqrt{x}+3 )}{(\sqrt{x}-3 )\times (\sqrt{x}+3 )}=\frac{\sqrt{x}-2 }{\sqrt{x}-3 }.}
Example 7
Compute the value of the expression \latex{\frac{\sqrt{x}+1 }{\sqrt{x}-2 }+\frac{\sqrt{x}-1 }{\sqrt{x}+2 }} for \latex{x=\frac{2}{3}.}
Solution
Obviously, \latex{x\geq 0} and \latex{x\neq 4.} First, find the common denominator:
\latex{\frac{\sqrt{x}+1 }{\sqrt{x}-2 }+\frac{\sqrt{x}-1 }{\sqrt{x}+2 }=\frac{(\sqrt{x}+1 )\times (\sqrt{x}+2 )+(\sqrt{x}-1 )\times (\sqrt{x}-2 )}{(\sqrt{x}-2 )\times (\sqrt{x}+2 )}=}
\latex{=\frac{x+2\times \sqrt{x}+\sqrt{x}+2+x-2\times \sqrt{x }-\sqrt{x}+2 }{x-4}=\frac{2x+4}{x-4}.}
\latex{=\frac{x+2\times \sqrt{x}+\sqrt{x}+2+x-2\times \sqrt{x }-\sqrt{x}+2 }{x-4}=\frac{2x+4}{x-4}.}
Substitute into this expression:
\latex{\frac{2\times \frac{2}{3}+4 }{\frac{2}{3}-4 }= \frac{\frac{16}{3} }{-\frac{10}{3} }=-\frac{16}{10}=-1.6.}
Do not be surprised by the result being negative, because for \latex{0\leq x\lt 1} both fractions are negative.
Example 8
Which one is greater:
- \latex{2\times \sqrt[4]{10}} or \latex{3\times \sqrt[4]{2};}
- \latex{\sqrt{5}\times \sqrt[3]{2}} or \latex{\sqrt{2}\times \sqrt[3]{5}?}
Solution (a)
In both cases, move everything under the radical sign:
\latex{2\times \sqrt[4]{10}=\sqrt[4]{2^{4} }\times \sqrt[4]{10}=\sqrt[4]{2^{4}\times 10 }=\sqrt[4]{160};}
\latex{3\times \sqrt[4]{2}=\sqrt[4]{3^{4} }\times \sqrt[4]{2}=\sqrt[4]{3^{4}\times 2 }=\sqrt[4]{162}.}
\latex{3\times \sqrt[4]{2}=\sqrt[4]{3^{4} }\times \sqrt[4]{2}=\sqrt[4]{3^{4}\times 2 }=\sqrt[4]{162}.}
Since the fourth root function is strictly increasing (Figure 2),
\latex{\sqrt[4]{162}\gt \sqrt[4]{160},} therefore \latex{2\times \sqrt[4]{10}\lt 3\times \sqrt[4]{2}. }
\latex{y=\sqrt[4]{x}}
\latex{ 1 }
\latex{ 1 }
\latex{ y }
\latex{ x }
Figure 2
Solution (b)
Write both expressions using only one radical sign:
\latex{\sqrt{5}\times \sqrt[3]{2}=\sqrt[6]{5^{3} }\times \sqrt[6]{2^{2} }=\sqrt[6]{5^{3}\times 2^{2} }=\sqrt[6]{500};}
\latex{\sqrt{2}\times \sqrt[3]{5}=\sqrt[6]{2^{3} }\times \sqrt[6]{5^{2} }=\sqrt[6]{2^{3}\times 5^{2} }=\sqrt[6]{200}.}
\latex{\sqrt{2}\times \sqrt[3]{5}=\sqrt[6]{2^{3} }\times \sqrt[6]{5^{2} }=\sqrt[6]{2^{3}\times 5^{2} }=\sqrt[6]{200}.}
Since the sixth root function is strictly increasing (Figure 3), it follows that
\latex{\sqrt[6]{500}\gt \sqrt[6]{200},} therefore \latex{\sqrt{5}\times \sqrt[3]{2}\gt \sqrt[3]{5}.}
\latex{y=\sqrt[6]{x}}
\latex{ 1 }
\latex{ x }
\latex{ 1 }
\latex{ y }
Figure 3
Example 9
Compute the following expressions:
- \latex{(\sqrt[5]{2}+3\times \sqrt[5]{8} )\times (\sqrt[5]{16}-\sqrt[5]{4} );}
- \latex{(\sqrt[3]{4}-2\times \sqrt[4]{8} )\times (2\times \sqrt[4]{2}-\sqrt[6]{2} ).}
Solution (a)
Expand the parentheses and remove everything possible from under the radical sign:
\latex{(\sqrt[5]{2}+3\times \sqrt[5]{8} )\times (\sqrt[5]{16}-\sqrt[5]{4} )=}
\latex{=\sqrt[5]{2}\times \sqrt[5]{16}-\sqrt[5]{2}\times \sqrt[5]{4}+3\times \sqrt[5]{8}\times \sqrt[5]{16}-3\times \sqrt[5]{8}\times \sqrt[5]{4}=}
\latex{=\sqrt[5]{32}-\sqrt[5]{8}+3\times \sqrt[5]{128}-3\times \sqrt[5]{32}=}
\latex{=2-\sqrt[5]{8}+3\times \sqrt[5]{32\times 4}-3\times 2=-4-\sqrt[5]{8}+6\times \sqrt[5]{4}.}
\latex{=\sqrt[5]{2}\times \sqrt[5]{16}-\sqrt[5]{2}\times \sqrt[5]{4}+3\times \sqrt[5]{8}\times \sqrt[5]{16}-3\times \sqrt[5]{8}\times \sqrt[5]{4}=}
\latex{=\sqrt[5]{32}-\sqrt[5]{8}+3\times \sqrt[5]{128}-3\times \sqrt[5]{32}=}
\latex{=2-\sqrt[5]{8}+3\times \sqrt[5]{32\times 4}-3\times 2=-4-\sqrt[5]{8}+6\times \sqrt[5]{4}.}
Solution (b)
After expanding the parentheses, transform such that the indices are equal where it is needed:
\latex{(\sqrt[3]{4}-2\times \sqrt[4]{8} )\times (2\times \sqrt[4]{2}+\sqrt[6]{2} )=}
\latex{=2\times \sqrt[3]{4}\times \sqrt[4]{2}+\sqrt[3]{4}\times \sqrt[6]{2}-2\times \sqrt[4]{8}\times 2\sqrt[4]{2}-2\times \sqrt[4]{8}\sqrt[6]{2}=}
\latex{=2\times \sqrt[12]{4^{2}\times 2^{3} }+\sqrt[6]{4^{2}\times 2 }-4\times \sqrt[4]{16}-2\times \sqrt[12]{8^{3}\times 2^{2} }=}
\latex{=2\times \sqrt[12]{2^{11} }+\sqrt[6]{2^{5} }-4\times 2-2\times \sqrt[12]{2^{11} }=\sqrt[6]{2^{5} }-8.}
\latex{=2\times \sqrt[3]{4}\times \sqrt[4]{2}+\sqrt[3]{4}\times \sqrt[6]{2}-2\times \sqrt[4]{8}\times 2\sqrt[4]{2}-2\times \sqrt[4]{8}\sqrt[6]{2}=}
\latex{=2\times \sqrt[12]{4^{2}\times 2^{3} }+\sqrt[6]{4^{2}\times 2 }-4\times \sqrt[4]{16}-2\times \sqrt[12]{8^{3}\times 2^{2} }=}
\latex{=2\times \sqrt[12]{2^{11} }+\sqrt[6]{2^{5} }-4\times 2-2\times \sqrt[12]{2^{11} }=\sqrt[6]{2^{5} }-8.}
Example 10
What powers of \latex{ 3 } are the following?
- \latex{\frac{3^{\frac{2}{5} }\times 9^{\frac{3}{5} }\times 27^{-\frac{1}{5} } }{81^{\frac{1}{5} } };}
- \latex{\frac{\sqrt{3}\times \sqrt[3]{9}\times \sqrt{27^{-1} } }{\sqrt[3]{81} };}
- \latex{\sqrt{3\times \sqrt[3]{3\times \sqrt[5]{3^{-1} } } }.}
Solution (a)
\latex{\frac{3^{\frac{2}{5} }\times 9^{\frac{3}{5} }\times 27^{-\frac{1}{5} } }{81^{\frac{1}{5} } }=\frac{3^{\frac{2}{5} }\times (3^{2} )^{\frac{3}{5} }\times (3^{3} )^{-\frac{1}{5} } }{(3^{4} )^{\frac{1}{5} } }= \frac{3^{\frac{2}{5} }\times 3^{\frac{6}{5} }\times 3^{-\frac{3}{5} } }{3^{\frac{4}{5} } }= \frac{3^{\frac{2}{5}+\frac{6}{5}+\frac{3}{5} } }{3^{\frac{4}{5} } }=\frac{3^{\frac{5}{5} } }{3^{\frac{4}{5} } }=3^{\frac{1}{5} }.}
Therefore the expression is the \latex{\frac{1}{5}}th power of \latex{ 3 }.
Solution (b)
\latex{\frac{\sqrt{3}\times \sqrt[3]{9}\times \sqrt{27^{-1} } }{\sqrt[3]{81} }=\frac{3^{\frac{1}{2} }\times 9^{\frac{1}{3} }\times (27^{-1} )^{\frac{1}{2} } }{81^{\frac{1}{3} } }= \frac{3^{\frac{1}{2} }\times 3^{\frac{2}{3} }\times 3 ^{-\frac{3}{2} } }{3^{\frac{4}{3} } }= \frac{3^{\frac{1}{2}+\frac{2}{3}+\frac{3}{2} } }{3^{\frac{4}{3} } }=\frac{3^{-\frac{1}{3} } }{3^{\frac{4}{3} } }=3^{-\frac{5}{3} }.}
Therefore the expression is the \latex{-\frac{5}{3}} power of \latex{ 3 }.
Solution (c)
Write the expression using only one radical sign and then transform into exponential form:
\latex{\sqrt{3\times \sqrt[3]{3\times \sqrt[5]{3^{-1} } } }=\sqrt{3\times \sqrt[3]{\sqrt[5]{3^{5}\times 3^{-1} } } }=\sqrt{3\times \sqrt[15]{3^{4} } }=}
\latex{=\sqrt{\sqrt[15]{3^{15}\times 3^{4} } }=\sqrt[30]{3^{19} }=3^{\frac{19}{30} }.}
\latex{=\sqrt{\sqrt[15]{3^{15}\times 3^{4} } }=\sqrt[30]{3^{19} }=3^{\frac{19}{30} }.}
Therefore the expression is the \latex{\frac{19}{30}} th power of \latex{ 3 }.
Logarithm
DEFINITION: The base a logarithm of a number b is the power to which \latex{ a } must be raised in order to produce \latex{ b }, where \latex{a\gt 0;a\neq 1} and \latex{b\gt 0.}
It is denoted by \latex{\log _{a}b.}
It is denoted by \latex{\log _{a}b.}
The base \latex{ 10 } logarithm is usually denoted by log.
The logarithm with base e is usually denoted by ln.
The logarithm with base e is usually denoted by ln.
THEOREM: Theorem. The logarithm of a product equals the sum of the logarithms of the multiplicands.
\latex{\log _{a}(x\times y)=\log _{a}x+\log _{a}y,}
where \latex{x,y\gt 0;a\gt 0;a\neq 1.}
where \latex{x,y\gt 0;a\gt 0;a\neq 1.}
THEOREM: The logarithm of a fraction equals the difference of the logarithms of the numerator and the denominator.
\latex{\log _{a}\frac{x}{y}=\log _{a}x-\log _{a}y,}
where \latex{x,y\gt 0;a\gt 0;a\neq 1.}
where \latex{x,y\gt 0;a\gt 0;a\neq 1.}
THEOREM: The logarithm of a power equals the product of the exponent and the logarithm of the base.
\latex{\log _{a}x^{k}=k\times \log _{a}x,}
where \latex{x\gt 0;a\gt 0;a\neq 1;k\in \R.}
where \latex{x\gt 0;a\gt 0;a\neq 1;k\in \R.}
THEOREM: \latex{\log _{a}b=\frac{\log _{c}b }{\log _{c}a },} where \latex{a,b,c\gt 0;a\neq 1;c\neq 1.}
Example 11
What is the greatest subset of the set of real numbers for which the following expressions are valid?
- \latex{\log _{5}(4-5x)-\log _{5}(2x+3);}
- \latex{\log (|x-1|-2)+2\times \log (4x-3).}
Solution (a)
The expression is valid if \latex{4-5x\gt 0} and \latex{2x+3\gt 0.} (Figure 4)
Their solutions are \latex{x\lt \frac{4}{5}} and \latex{x\gt -\frac{3}{2}.}
The common solution is \latex{-\frac{3}{2}\lt x\lt \frac{4}{5}.}
Their solutions are \latex{x\lt \frac{4}{5}} and \latex{x\gt -\frac{3}{2}.}
The common solution is \latex{-\frac{3}{2}\lt x\lt \frac{4}{5}.}
\latex{ 0 }
\latex{ \frac{4}{5} }
\latex{ -\frac{3}{2} }
Figure 4
Solution (b)
The expression is valid if \latex{|x-1|-2\gt 0} and \latex{4x-3\gt 0.}
The solution of the first one is \latex{|x-1|\gt 2.}
If \latex{x\geq 1:}
The solution of the first one is \latex{|x-1|\gt 2.}
If \latex{x\geq 1:}
\latex{\begin{rcases}x-1\gt 2 \\ x\gt 3\end{rcases}} the common solution is \latex{x\gt 3.}
If \latex{x\lt 1:}
\latex{\begin{rcases}-x+1\gt 2 \\ x\lt -1\end{rcases}} the common solution is \latex{x\lt -1.}
The solution of the second inequality is \latex{x\gt \frac{3}{4}.}
The expression is valid if both condition is satisfied, that is, if \latex{x\gt 3.} (Figure 5)
The expression is valid if both condition is satisfied, that is, if \latex{x\gt 3.} (Figure 5)
\latex{ 0 }
\latex{ \frac {3}{4} }
\latex{ -1 }
\latex{ 3 }
Figure 5
Example 12
Determine the value of the following exponentiations:
- \latex{10^{2-\log 8};}
- \latex{8^{\log _{8}5-1 };}
- \latex{0,25^{\log _{2}5 };}
- \latex{3^{3-\log _{9}49 }.}
Solution (a)
Apply the identities of exponentiation and the definition of logarithm:
\latex{10^{2-\log _{8} }=\frac{10^{2} }{10^{\log8 } }=\frac{100}{8}=12.5.}
Solution (b)
Using the same method,
\latex{8^{\log _{8}5-1 }=\frac{8^{\log _{8}5 } }{8}=\frac{5}{8}.}
Solution (c)
Express the base as a power of \latex{ 2 } and apply the previous method:
\latex{0,25^{\log _{2}5 }=(2^{-2} )^{\log_{2}5 }=(2^{\log _{2}5 } )^{-2}=5^{-2}=\frac{1}{25}.}
Solution (d)
In order to apply the definition, first express the base of the exponential as a power of the base of the logarithm:
\latex{3^{3-\log _{9}49 }=\frac{3^{3} }{3^{\log _{9}49 } }=\frac{27}{ (9^{\frac{1}{2} } )^{\log _{9}49 } }=\frac{27}{(9^{\log _{9}49 } )^{\frac{1}{2} } }=\frac{27}{49^{\frac{1}{2} } }=\frac{27}{1}.}
Example 13
Compute the following expressions:
- \latex{\log _{2}\frac{4\times 2^{3}\times 16 }{64^{2} };}
- In \latex{\frac{e^{-\frac{1}{3} }\times \sqrt{e}\times \sqrt[3]{e^{2} }\times e^{2} }{\sqrt[6]{e^{5} }\times e^{-1} }. }
Solution (a)
Method I: Express the numbers as powers of \latex{ 2 } and compute the exponentiations:
\latex{\log^{2}\frac{4\times 2^{3}\times 16 }{64^{2} }=\log _{2}\frac{2^{2}\times 2^{3}\times 2^{4} }{(2^{6} )^{2} }=\log_{2}\frac{2^{9} }{2^{12} }=\log _{2}2^{-3}=-3.}
Method II: Apply the identities of logarithm:
\latex{\log _{2}\frac{4\times 2_{3}\times 16 }{64^{2} }=\log _{2}4+\log _{2}2^{3}+\log _{2}16-\log _{2}64^{2}=}
\latex{=2+3+4-2\times 6=-3.}
\latex{=2+3+4-2\times 6=-3.}
Solution (b)
Express every term as a power of \latex{ e }:
\latex{In\frac{e^{-\frac{1}{3} }\times \sqrt{e}\times \sqrt[3]{e^{2} }\times e^{2} }{\sqrt[6]{e^{5} }\times e^{-1} }= In \frac{e^{\frac{1}{3}+\frac{1}{2}+\frac{2}{3}+2 } }{e^{\frac{5}{6}-1 } } =In \frac{e^{\frac{17}{6} } }{e^{-\frac{1}{6} } }=In e^{3}=3.}
Exercises
{{exercise_number}}. What is the greatest power of \latex{ 3 } which is a divisor of \latex{15^{4}\times 27^{5}\times 45^{3}?}
{{exercise_number}}. How many zero digits are there at the end of the product \latex{125^{5}\times 144^{3}\times 96^{2}?}
{{exercise_number}}. The heart of an adult performs work approximately enough to lift a weight of \latex{ 1\, kg } to \latex{ 20\, cm } for each heartbeat. Our heart beats approximately \latex{ 70 } times per minute.
- How high could we lift our body using all the work done by our heart during our life so far?
- How large an object can be lifted \latex{ 1\,m } high using this force?
{{exercise_number}}. Compute the following fractions:
- \latex{\frac{16^{-3}\times 64^{4}\times 2^{3} }{32^{2}\times 8^{-2} };}
- \latex{\frac{36^{-4}\times 54^{-2} }{108^{-3} };}
- \latex{\frac{16^{-\frac{1}{4} }\times 8^{\frac{2}{3} } }{4^{\frac{3}{2} }\times 32^{-\frac{1}{5} } }.}
{{exercise_number}}. Prove that
- \latex{\sqrt{9-4\sqrt{5} }=\sqrt{5}-2;}
- \latex{\sqrt{16-6\sqrt{7} }=3-\sqrt{7}.}
{{exercise_number}}. Which one is larger:
- \latex{\frac{15}{\sqrt{7}-\sqrt{2} }} or \latex{\frac{1}{\sqrt{19}-3\times \sqrt{2} }};
- \latex{\frac{16}{\sqrt{15}+\sqrt{7} }} or \latex{\frac{12}{\sqrt{7}+\sqrt{13} }?}
{{exercise_number}}. Determine the largest subset of the set of real numbers for which the following expressions are valid. Simplify the expressions.
- \latex{\frac{\sqrt{a-3} }{\sqrt{5a+25} }\times \frac{\sqrt{a+5} }{\sqrt{2a-6} };}
- \latex{\frac{3\times \sqrt{b}-12 }{2\times\sqrt{b}-2 }\div \frac{\sqrt{b}-4 }{4\times \sqrt{b}-4 }.}
{{exercise_number}}. Prove that the following expression is divisible by \latex{ 4 } for every positive integer \latex{ n }:
\latex{\sqrt{\frac{2n+\sqrt{4n^{2}-1 } }{2n-\sqrt{4n^{2}-1 } } }+\sqrt{\frac{2n-\sqrt{4n^{2}-1 } }{2n+\sqrt{4n^{2}-1 } } }.}
{{exercise_number}}. Compute the following:
- \latex{(\sqrt[3]{7}-\sqrt[3]{3} )\times (\sqrt[3]{49}+\sqrt[3]{21}+\sqrt[3]{9} );}
- \latex{(\sqrt[3]{10}-\sqrt[3]{6} )\times (\sqrt[3]{100}-\sqrt[3]{60}+\sqrt[3]{36} );}
- \latex{(\sqrt[4]{11}-\sqrt[4]{5} )\times (\sqrt{11}+\sqrt{5} )\times (\sqrt[4]{11}+\sqrt[4]{5} ).}
{{exercise_number}}. Put the following numbers into an increasing order:
- \latex{27^{\frac{1}{3} },\left(\frac{1}{3} \right)^{-2},9^{\frac{3}{2} },\left(\frac{81}{4} \right)^{-\frac{1}{2} },\left(\frac{3}{5} \right)^{-2},\left(\frac{5}{3} \right)^{-1};}
- \latex{7^{\log _{7}5-1 },7^{1-\log _{49}25 },\left(\frac{1}{7} \right)^{\log _{7}3 },7^{\log _{\frac{1}{7} }5 },49^{\log _{7}2 },7^{\log _{13}1 };}
- \latex{\log _{3}\frac{1}{27},\log _{25}5,\log _{27}\frac{1}{3},\log _{2}0,125.}
{{exercise_number}}. Find the value of \latex{ x } if
- \latex{\log x=\log 20-\log 15+\log 30-\log 4;}
- \latex{\log x=2\times \log 5-3\times \log 2;}
- \latex{\log x=\frac{1}{2}\times \log 16-\frac{1}{3}\times \log 37-2\times \log 2+\frac{1}{4}\times \log 81.}