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Mathematics 11.

Table of contents
Combinatorics, graphs
Powers, roots, logarithms
1. Exponentiation and computing the roots (reminder)
2. Power functions and root functions
3. Power with fractional (rational) exponent
4. Power with irrational exponent, exponential function
5. Exponential equations, simultaneous exponential equations and exponential inequalities
6. The concept of the logarithm
7. The logarithm function
8. The logarithmic identities
9. Logarithmic equations, simultaneous logarithmic equations and logarithmic inequalities
10. Practical applications
Applications of trigonometry
Functions
Coordinate geometry
Calculation of probability, statistics
Power with fractional (rational) exponent
We want to extend the concept of power to all rational indices so that the identities of integer indices (which make it easier to calculate with powers) will still stay true. This requirement is called the principle of constancy (permanent \latex{=} continuance, continuity).
The method is similar to when extending the concept of exponentiation to the integer indices.
The operational properties of powers and roots show a close relationship.
principle of constancy
We can raise a product to a power factor by factor and we can take the root of a product factor by factor too:
\latex{\left( a \times b\right)^n=a^n \times b^n} and \latex{\sqrt[n]{a \times b}=\sqrt[n]{a} \times \sqrt[n]{b}}.
A similar statement is also true for the operations with fractions:
\latex{\left( \frac{a}{b}\right)^n=\frac{a^n}{b^n}} and \latex{\sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}}.
The operations of exponentiation and taking the root are interchangeable:
\latex{\sqrt[n]{k}=\left(\sqrt[n]{a}\right)^k}.
The relation between the two operations makes it possible to extend the exponentiation to fractional exponents.
Let us use the identity \latex{V} of exponentiation and the definition of roots:
\latex{\begin{rcases}\left(7^{\frac{2}{3}}\right)^3=7^2\\\\\left(\sqrt[3]{7^2}\right)^3=7^2\end{rcases}}so let \latex{7^{\frac{2}{3}}=\sqrt[3]{7^2}}.
The formal application
of identity \latex{V} for
\latex{n=\frac{2}{3}}:
Based on the
definition of roots:
\latex{7^{\frac{2}{3}}=\sqrt[3]{7^2}}
DEFINITION: The \latex{\frac{m}{n}}th power of the positive number \latex{a} is the \latex{n^{\text{th}}} root of the \latex{m^{\text{th}}} power of the base \latex{a}:
\latex{a^{\frac{m}{n}}=\sqrt[n]{a^m}}, where \latex{a \gt 0}; \latex{m \in \Z}; \latex{n \in \N} and \latex{n \geq 2}.
power with fractional exponent
\latex{a^{\frac{m}{n}}=\sqrt[n]{a^m}}
\latex{a \gt 0}; \latex{m \in \Z}; \latex{n \in \N}; \latex{n \geq 2}
Notes:
  1. Every non-integer rational number can be expressed in a form corresponding to the conditions. For example:
\latex{-\frac{7}{3}=\frac{-7}{3}}, \latex{0.23=\frac{23}{100}}, \latex{-1.41=\frac{-141}{100}}.
  1. If we want to extend the concept of exponentiation to all rational indices, then \latex{a \leq 0} cannot occur, since for example the expressions \latex{\left( -5 \right)^{\frac{1}{2}}=\sqrt{-5}} or \latex{0^{-\frac{1}{3}}=\frac{1}{\sqrt[3]{0}}} cannot be defined on the set of real numbers.
◆ ◆ ◆
Every rational number can be expressed as the quotient of two whole numbers in infinitely many ways, for example:
\latex{-\frac{7}{3}=\frac{-7}{3}=\frac{-14}{6}=\frac{-21}{9}=\frac{-700}{300}=...}
Is it true that we get the same result when raising a positive number to any form of the rational exponent?
THEOREM: If \latex{a\gt 0}; \latex{k}, \latex{l}, \latex{m}, \latex{n \in \Z}; \latex{n}, \latex{l \geq 2} and \latex{\frac{m}{n}=\frac{k}{l}}, then \latex{a^{\frac{m}{n}}=a^\frac{k}{l}}.
NICOLE ORESMICUS
(\latex{ 1323–1382 }), he first wrote about the  generalisation of
Proof
If \latex{\frac{m}{n}=\frac{k}{l}}, then \latex{m \times l = n \times k}.
\latex{a^{\frac{m}{n}}=\sqrt[n]{a^m}=\sqrt[n\times l]{a^{m \times l}}=\sqrt[n\times l]{a^{n \times k}}=\sqrt[l]{a^k}=a^{\frac{k}{l}}}
\latex{\huge\vert}
definition
\latex{\huge\vert}
identity E)
\latex{\huge\vert}
\latex{m \times l = n \times k}
\latex{\huge\vert}
identity E)
\latex{\huge\vert}
definition
powers to positive fractional exponents in his work On Ratios of Ratios. In one of his later works he also described some physical and geometrical applications of powers with fractional
Example 1
Let us calculate the values of the following powers:
exponents. For example, with his notation:
  1. \latex{1^{-\frac{3}{4}};}
  1. \latex{8^{\frac{2}{3}};}
  1. \latex{9^{\frac{3}{2}};}
  1. \latex{27^{-\frac{4}{3}};}
  1. \latex{0.0625^{-\frac{3}{4}}.}
\latex{8^{\frac{2}{3}}=\frac{2}{3} \frac{p}{8}}.
Solution
  1. \latex{1^{-\frac{3}{4}}=\sqrt[4]{1^{-3}}=\sqrt[4]{\frac{1}{1^3}}=\sqrt[4]{1}=1}
  2. \latex{8^{\frac{2}{3}}=\sqrt[3]{8^2}=\sqrt[3]{64}=4}
  3. \latex{9^{\frac{3}{2}}=\sqrt{9^3}=\sqrt{\left( 3^2\right)^3}=\sqrt{3^6}=27}
  4. \latex{27^{-\frac{4}{3}}=\sqrt[3]{27^{-4}}=\sqrt[3]{\frac{1}{27^4}}=\frac{1}{\sqrt[3]{\left(3^3\right)^4}}=\frac{1}{\sqrt[3]{3^{12}}}=\frac{1}{3^4}=\frac{1}{81}}
  5. \latex{0.0625^{-\frac{3}{4}}=\sqrt[4]{\left( 2^{-4}\right)^{-3}}=\sqrt[4]{2^{12}}=2^3=8}
\latex{0.0625=\frac{625}{10,000}=\frac{1}{16}=2^{-4}}
Example 2
Let us express the following powers with only root signs:
  1. \latex{5^{\frac{1}{5}}};
  1. \latex{7^{-\frac{4}{10}}};
  1. \latex{\left(\frac{4}{5}\right)^{\frac{3}{4}}};
  1. \latex{\left(\frac{1}{3}\right)^{-\frac{2}{3}}}.
Solution
According to the definition:
  1. \latex{5^{\frac{1}{5}}=\sqrt[5]{5}};
  1. \latex{7^{-\frac{4}{10}}=7^{-\frac{2}{5}}=\sqrt[5]{7^{-2}}=\sqrt[5]{\frac{1}{49}}};
  1. \latex{\left(\frac{4}{5}\right)^{\frac{3}{4}}=\sqrt[4]{\left(\frac{4}{5}\right)^3}=\sqrt[4]{\frac{64}{125}}};
  1. \latex{\left(\frac{1}{3}\right)^{-\frac{2}{3}}=\sqrt[3]{\left(\frac{1}{3}\right)^{-2}}=\sqrt[3]{9}}.
\latex{\left( \frac{1}{3}\right)^{-2}=\left(3^{-1}\right)^{-2}=3^2=9}
◆ ◆ ◆
The following question arises: what is the value of the expression mentioned in the introduction?
Based on the definition, calculated by a pocket calculator and rounded to three decimal places:
\latex{7^{\frac{2}{3}}=\sqrt[3]{7^2}=\sqrt[3]{49}\approx3.659}
Since we introduced the powers with fractional exponent with the help of roots, there are only “a few” powers the exact values of which can be calculated; often we can only give approximate values.
◆ ◆ ◆
The power with fractional exponent was defined by applying the identity \latex{V} of powers. We are going to show using some examples that all the identities stay true.
  1. If \latex{a \gt 0}:
\latex{a^{\frac{2}{3}}\times a^{\frac{1}{4}}=a^{\frac{8}{12}}\times a^{\frac{3}{12}}=\sqrt[12]{a^8} \times \sqrt[12]{a^3}=\sqrt[12]{a^8 \times a^3}=\sqrt[12]{a^{11}}=a^{\frac{11}{12}}=a^{\frac{2}{3}+\frac{1}{4}}}
\latex{\huge\vert}
definition
\latex{\huge\vert}
identity E)
\latex{\huge\quad\vert}
expanding fractions
\latex{\huge\vert}
identity I
for integer exponent
\latex{\huge\vert\quad}
definition
  1. If \latex{a \gt 0}:
\latex{\frac{a^{\frac{1}{4}}}{a^{\frac{1}{6}}}=\frac{a^{\frac{3}{12}}}{a^{\frac{2}{12}}}=\frac{\sqrt[12]{a^3}}{\sqrt[12]{a^2}}=\sqrt[12]{\frac{a^3}{a^2}}=\sqrt[12]{a}=a^{\frac{1}{12}}=a^{\frac{1}{4}-\frac{1}{6}}}.
\latex{\Large\vert}
definition
\latex{\Huge\text{\textbar}}
identity B)
\latex{\Huge\quad\vert}
expanding fractions
\latex{\vert}
identity II
for integer
exponent
\latex{\huge\backslash\quad}
definition
  1. If \latex{a,b \gt 0}:
\latex{\left( a \times b \right)^{\frac{7}{5}}=\sqrt[5]{\left( a \times b \right)^{7}}=\sqrt[5]{a^7 \times b^7}=\sqrt[5]{a^7}\times\sqrt[5]{b^7}=a^{\frac{7}{5}}\times b^{\frac{7}{5}}}.
\latex{\huge\vert}
definition
\latex{\huge\vert}
identity A)
\latex{\huge\vert}
identity III
for integer exponent
\latex{\huge\vert}
definition
  1. If \latex{a,b \gt 0}:
\latex{\left({\frac{a}{b}}\right)^{\frac{5}{7}}=\sqrt[7]{\left(\frac{a}{b}\right)^5}=\sqrt[7]{\frac{a^5}{b^5}}=\frac{\sqrt[7]{a^5}}{\sqrt[7]{b^5}}=\frac{a^{\frac{5}{7}}}{b^{\frac{5}{7}}}}.
\latex{\huge\vert}
definition
\latex{\huge\vert\quad}
identity B)
\latex{\huge\vert}
identity IV
for integer exponent
\latex{\huge\backslash\quad}
definition
MICHAEL STÍFEL
(\latex{ 1487–1567 }), German mathematician first mentions the powers of \latex{2} with zero and
  1. If both exponents are fractions, \latex{a \gt 0}:
\latex{\left(a^{\frac{5}{7}}\right)^{\frac{2}{3}}=\sqrt[3]{\left(a^{\frac{5}{7}}\right)^2}=\sqrt[3]{\left( \sqrt[7]{a^5}\right)^2}=\sqrt[3]{\sqrt[7]{a^{5 \times 2}}}=\sqrt[3 \times 7]{a^{5\times 2}}=a^{\frac{5 \times 2}{3 \times 7}}}.
\latex{\huge\vert}
definition
\latex{\huge\vert}
identity C)
\latex{\huge\vert}
definition
\latex{\huge\vert}
definition
\latex{\huge\vert}
identity D)
As we can see the definition of the power with fractional exponent satisfies the requirements of the principle of constancy; the identities of powers stay true after the extension too.
negative integer exponents besides the positive integer exponents in his work Arithmetica integra.
Example 3
Let us calculate the values of the expressions:
  1. \latex{\frac{7^{\frac{2}{3}}\times7^{\frac{1}{6}} \times 7^{\frac{3}{4}}}{7^{\frac{1}{12}} \times 49^{\frac{1}{4}}}};
  1. \latex{\left( \frac{2}{3} \right)^{\frac{1}{4}} \times \frac{6^{\frac{1}{5}} \times 3^{\frac{1}{2}}}{4^{\frac{1}{10}} \times9^{\frac{1}{5}}}};
  1. \latex{\frac{10^{\frac{1}{8}} \times 25^{\frac{1}{3}} \times 4^{\frac{1}{4}}}{100^{\frac{3}{8}}}}.
Solution
  1. \latex{\frac{7^{\frac{2}{3}}\times 7^{\frac{1}{6}} \times 7^{\frac{3}{4}}}{7^{\frac{1}{12}} \times 49^{\frac{1}{4}}}=\frac{7^{\frac{2}{3}+\frac{1}{6}+\frac{3}{4}}}{7^{\frac{1}{12}} \times \left(7^2\right)^{\frac{1}{4}}}=\frac{7^{\frac{8+2+9}{12}}}{7^{\frac{1}{12}} \times 7^{\frac{2}{4}}}=\frac{7^{\frac{19}{12}}}{7^{\frac{1+6}{12}}}=7^{\frac{19}{12}-\frac{7}{12}}=7}
  2. \latex{\left( \frac{2}{3} \right)^{\frac{1}{4}} \times \frac{6^{\frac{1}{5}} \times 3^{\frac{1}{2}}}{4^{\frac{1}{10}} \times 9^{\frac{1}{5}}}=\frac{2^{\frac{1}{4}}\times \left(2 \times 3 \right)^{\frac{1}{5}}\times3^{\frac{1}{2}}}{3^{\frac{1}{4}}\times\left(2^2 \right)^{\frac{1}{10}}\times\left( 3^2\right)^{\frac{1}{5}}}=\frac{2^{\frac{1}{4}} \times 2^{\frac{1}{5}}\times 3^{\frac{1}{5}}\times 3^{\frac{1}{2}}}{3^{\frac{1}{4}}\times 2^{\frac{2}{10}}\times 3^{\frac{2}{5}}}=\frac{2^{\frac{1}{4}}\times2^{\frac{1}{5}}}{2^{\frac{1}{5}}}\times\frac{3^{\frac{1}{5}+\frac{1}{2}}}{3^{\frac{1}{4}+\frac{2}{5}}}=2^{\frac{1}{4}}\times\frac{3^{\frac{7}{10}}}{3^{\frac{13}{20}}}=}\latex{=2^{\frac{1}{4}}\times3^{\frac{7}{10}-\frac{13}{20}}=2^{\frac{1}{4}}\times3^{\frac{1}{20}}}
  3. \latex{\frac{10^{\frac{1}{8}} \times 25^{\frac{1}{3}}\times4^{\frac{1}{4}}}{100^{\frac{3}{8}}}=\frac{\left( 2 \times 5\right)^{\frac{1}{8}}\times\left( 5^2\right)^{\frac{1}{3}}\times \left(2^2\right)^{\frac{1}{4}}}{\left( 2^2 \times 5^2\right)^{\frac{3}{8}}}=\frac{2^{\frac{1}{8}}\times 5^{\frac{1}{8}}\times 5^{\frac{3}{2}} \times 2^{\frac{2}{4}}}{2^{\frac{6}{8}}\times 5^{\frac{6}{8}}}=\frac{2^{\frac{1}{8}+\frac{1}{2}}}{2^{\frac{6}{8}}}\times\frac{5^{\frac{1}{8}+\frac{2}{3}}}{5^{\frac{6}{8}}}=\frac{2^{\frac{5}{8}}}{2^{\frac{6}{8}}}\times\frac{5^{\frac{19}{24}}}{5^{\frac{6}{8}}}=}\latex{=2^{-\frac{1}{8}}\times 5^{\frac{1}{24}}}
Example 4
Let us rewrite the following expressions as powers of \latex{5}:
  1. \latex{\sqrt[3]{\sqrt[4]{5}}};
  1. \latex{\sqrt[5]{5^3}\times\sqrt[4]{5^3}};
  1. \latex{\frac{\sqrt[4]{125}\times \sqrt[3]{25} \times \sqrt[6]{125^5}}{\sqrt[3]{5^7}\times \sqrt[3]{625}}.}
Solution
  1. \latex{\sqrt[3]{\sqrt[4]{5}}}
  1. \latex{\sqrt[5]{5^3}\times\sqrt[4]{5^3}}
  1. \latex{\frac{\sqrt[4]{125}\times \sqrt[3]{25} \times \sqrt[6]{125^5}}{\sqrt[3]{5^7}\times \sqrt[3]{625}}}
Solution
  1. \latex{\sqrt[3]{\sqrt[4]{5}}=\left( 5^{\frac{1}{4}}\right)^{\frac{1}{3}}=5^{\frac{1}{12}}}
  2. \latex{\sqrt[5]{5^3}\times\sqrt[4]{5^3}=5^{\frac{3}{5}}\times 5^{\frac{3}{4}}=5^{\frac{3}{5}+\frac{3}{4}}=5^{\frac{27}{20}}}
  3. \latex{\frac{\sqrt[4]{125}\times \sqrt[3]{25} \times \sqrt[6]{125^5}}{\sqrt[3]{5^7}\times \sqrt[3]{625}}=\frac{\sqrt[4]{5^3}\times\sqrt[3]{5^2}\times\sqrt[6]{5^{15}}}{\sqrt[3]{5^7}\times\sqrt[3]{5^4}}=\frac{5^{\frac{3}{4}} \times 5^{\frac{2}{3}} \times 5^{\frac{15}{6}}}{5^{\frac{7}{3}} \times 5^{\frac{4}{3}}}=\frac{5^{\frac{3}{4}+\frac{2}{3}+\frac{15}{6}}}{5^{\frac{7}{3}+\frac{4}{3}}}=}\latex{\frac{5^{\frac{47}{12}}}{5^{\frac{11}{3}}}=5^{\frac{47}{12}-\frac{11}{3}}}=\latex{= 5^{\frac{3}{12}}= 5^{\frac{1}{4}}}
\latex{125^5=\left(5^3\right)^5=5^{15}}
We defined the powers with fractional exponent based on the roots.
In the case of specific calculation examples we can use the identities of both powers and roots.
Example 5
Let us simplify the following expressions:
  1. \latex{\frac{\sqrt[4]{a^3} \times a^{-\frac{1}{2}}}{a^{\frac{1}{4}} \times \sqrt[8]{a^5}}},  \latex{a \gt 0};
  1. \latex{\frac{\sqrt[3]{a}\times b^{\frac{2}{3}}\times \sqrt[4]{a^3 \times b}}{\sqrt{a} \times \left( a \times b^3 \right)^{\frac{3}{4}}}},  \latex{a}, \latex{b \gt 0};
  1. \latex{\frac{\sqrt[4]{b^{-\frac{2}{3}}} \times \sqrt[3]{a^{-\frac{1}{4}}} \times \left( a^{\frac{1}{2}} \right)^{\frac{1}{4}}}{\left( b^{-3} \right)^{\frac{1}{8}} \times \sqrt[4]{a^{-3}}\times \sqrt{\left(a \times b\right)^{-1}}}},  \latex{a}, \latex{b \gt 0}.
Solution (a)
Method I (with powers):
\latex{\frac{\sqrt[4]{a^3}\times a^{-\frac{1}{2}}}{a^{\frac{1}{4}} \times \sqrt[8]{a^5}}=\frac{a^{\frac{3}{4}}\times a^{-\frac{1}{2}}}{a^{\frac{1}{2}}\times a^{\frac{5}{8}}}=\frac{a^{\frac{3}{4}-\frac{1}{2}}}{a^{\frac{1}{4}+\frac{5}{8}}}=\frac{a^{\frac{1}{4}}}{a^{\frac{7}{8}}}=a^{\frac{1}{4}-\frac{7}{8}}=a^{-\frac{5}{8}}=\sqrt[8]{a^{-5}}}.
Method II (with roots):
\latex{\frac{\sqrt[4]{a^3}\times a^{-\frac{1}{2}}}{a^{\frac{1}{4}} \times \sqrt[8]{a^5}}=\frac{\sqrt[4]{a^3}\times \sqrt{a^{-1}}}{\sqrt[4]{a}\times\sqrt[8]{a^5}}=\frac{\sqrt[a]{a^6}\times \sqrt[8]{a^{-4}}}{\sqrt[8]{a^2}\times\sqrt[8]{a^5}}=\sqrt[8]{\frac{a^6 \times a^{-4}}{a^2 \times a^5}}=\sqrt[8]{\frac{1}{a^5}}=\sqrt[8]{a^{-5}}=a^{-\frac{5}{8}}}.
Solution (b)
\latex{\frac{\sqrt[3]{a}\times b^{\frac{2}{3}}\times \sqrt[4]{a^3 \times b}}{\sqrt{a} \times \left( a \times b^3 \right)^{\frac{3}{4}}}=\frac{a^{\frac{1}{3}}\times b^{\frac{2}{3}}\times\left( a^3 \times b\right)^{\frac{1}{4}}}{a^{\frac{1}{2}}\times\left( a \times b^3\right)^{\frac{3}{4}}}=\frac{a^{\frac{1}{3}}\times b^{\frac{2}{3}}\times a^{\frac{3}{4}}\times b^{\frac{1}{4}}}{a^{\frac{1}{2}}\times\left( a \times b^3\right)^{\frac{3}{4}}}=\frac{a^{\frac{1}{3} + \frac{3}{4}}\times b^{\frac{2}{3}+\frac{1}{4}}}{a^{\frac{1}{2}+\frac{3}{4}}\times b^{\frac{9}{4}}}=} \latex{=\frac{a^{\frac{13}{12}}\times b^{\frac{11}{12}}}{a^{\frac{5}{4}}\times b^{\frac{9}{4}}}=a^{\frac{13}{12}-\frac{5}{4}}\times b^{\frac{11}{12}-\frac{9}{4}}=a^{-\frac{2}{12}}\times b^{-\frac{16}{12}}=a^{-\frac{1}{6}}\times b^{-\frac{4}{3}}=\sqrt[6]{a^{-1}} \times \sqrt[3]{b^{-4}}}
Solution (c)
\latex{\frac{\sqrt[4]{b^{-\frac{2}{3}}}\times \sqrt[3]{a^{-\frac{1}{4}}}\times \left( a^{\frac{1}{2}}\right)^{\frac{1}{4}}}{\left( b^{-3} \right)^{\frac{1}{8}}\times \sqrt[4]{a^{-3}}\times \sqrt{\left( a \times b \right)^{-1}}}=\frac{b^{-\frac{2}{12}}\times a^{-\frac{1}{12}} \times a^{\frac{1}{8}}}{b^{-\frac{3}{8}}\times a^{-\frac{3}{4}}\times a^{-\frac{1}{2}} \times b^{-\frac{1}{2}}}=\frac{a^{-\frac{1}{12}+\frac{1}{8}}}{a^{-\frac{3}{4}-\frac{1}{2}}}\times \frac{b^{-\frac{1}{6}}}{b^{-\frac{3}{8}-\frac{1}{2}}}=} \latex{=a^{-\frac{1}{12}+\frac{1}{8}+\frac{3}{4}+\frac{1}{2}}\times b^{-\frac{1}{6}+\frac{3}{8}+\frac{1}{2}}=a^{\frac{31}{24}}\times b^{\frac{17}{24}}=\sqrt[24]{a^{31}\times b^{17}}}
Exercises
{{exercise_number}}. Express the following powers with root signs:
  1. \latex{3^{\frac{1}{5}}};
  1. \latex{2^{\frac{3}{4}}};
  1. \latex{4^{-\frac{5}{3}}};
  1. \latex{\left(\frac{3}{5}\right)^{\frac{2}{5}}};
  1. \latex{\left( \frac{2}{3}\right)^{-\frac{3}{7}}};
  1. \latex{\left(- \frac{3}{7}\right)^{\frac{7}{2}}}.
{{exercise_number}}. Rewrite the following expressions as powers of \latex{ 2 }:
  1. \latex{\sqrt[3]{2}};
  1. \latex{\sqrt[5]{2^3}};
  1. \latex{\sqrt[7]{\frac{1}{2^3}}};
  1. \latex{\sqrt[4]{\sqrt[3]{2^7}}};
  1. \latex{\sqrt[3]{2^2}\times\sqrt[4]{2^3}};
  1. \latex{\frac{\sqrt{2}\times\sqrt[6]{2^5}\times\sqrt[3]{2^3}}{\sqrt{8}\times\sqrt[3]{16}}};
  1. \latex{\sqrt{2 \times \sqrt[5]{2^2} \times \sqrt{2}}\times \sqrt[5]{4}.}
{{exercise_number}}. Rewrite the following expressions as powers of \latex{a}, \latex{b}, \latex{c}, \latex{d} where \latex{a}, \latex{b}, \latex{c}, \latex{d \gt 0}:
  1. \latex{\frac{\sqrt{a}\times a^{\frac{2}{3}}}{\sqrt[3]{a}}};
  1. \latex{\frac{\sqrt[4]{b^3}\times b^{-\frac{1}{2}}\times b^{\frac{5}{8}}}{\sqrt{\sqrt[4]{b}}\times \sqrt{b^{-3}}}};
  1. \latex{\frac{\sqrt[3]{c^2}\times c^{-\frac{3}{4}}\times c^{\frac{1}{6}}}{\sqrt{c^{-1}}\times \sqrt[6]{c^{-2}}}};
  1. \latex{\frac{d^{-\frac{2}{3}}\times d^{\frac{3}{4}}\times \sqrt[5]{d^2}}{\sqrt{d^{-3}}\times \sqrt[4]{d}\times d^{\frac{5}{6}}}}.
{{exercise_number}}. Sort the following numbers in ascending order:
  1. \latex{\sqrt[5]{3}}, \latex{3^{-\frac{2}{3}}}, \latex{\sqrt[4]{3}}, \latex{3^{\frac{1}{5}}}, \latex{3^{\frac{2}{3}}}, \latex{3^{-\frac{3}{2}}};
  1. \latex{4^{-\frac{1}{2}}}, \latex{\sqrt[7]{4}}, \latex{4^{-\frac{3}{2}}}, \latex{\sqrt[5]{16}}, \latex{2^{-1}}, \latex{2^{\frac{2}{7}}}.
Puzzle
Which number is greater: \latex{3^{\frac{3}{2}}} or \latex{5^{\frac{2}{3}}}?

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