 maths > exponents

root : An inverse of Exponent

what you'll learn...

overview

One of the inverses of exponent is root. Root is introduced with the following two.

•  first principle -- Root of a number to a given power of root is the base of the exponent with the given power.

•  Simplified Procedure -- Root of a number is found from prime-factorization of the numbers (if root evaluates to an integer).

inverses (recap)

If the sum and the first addend are given, then

If the sum and the second addend are given, then

subtraction is the inverse for both because addition is commutative
$3+2=5$$3 + 2 = 5$ and $2+3=5$$2 + 3 = 5$.

The inverse of multiplication is "division"

Multiplication is
multiplicand $×$$\times$ multiplier $=$$=$ product.

If the multiplicand and the product are given, then
multiplier $=$$=$ product $÷$$\div$ multiplicand

If the multiplier and the product are given, then
multiplicand $=$$=$ product $÷$$\div$ multiplier

Division serves as the inverse of multiplication for both multiplier and multiplicand.

Division is the inverse for both because, multiplication is commutative.
$3×2=6$$3 \times 2 = 6$ and $2×3=6$$2 \times 3 = 6$.

two inverses of exponent

What could be the inverse(s) of "exponent"?

Is it

(1) given result of exponentiation and base, find the power or

(2) given result of exponentiation and power, find the base

It is "both the above"

Two inverses are defined for exponents.

The exponent is not commutative, and that results in two inverses to exponent.
${3}^{2}=9$${3}^{2} = 9$ and ${2}^{3}=8$${2}^{3} = 8$
${a}^{b}\ne {b}^{a}$${a}^{b} \ne {b}^{a}$

Exponent is
${\left(\text{base}\right)}^{\text{power}}=\text{exp. result}$${\left(\textrm{b a s e}\right)}^{\textrm{p o w e r}} = \textrm{\exp . r e s \underline{t}}$

If exponentiation result and power are given, then
$\text{base}=\sqrt[\text{power}]{\text{result}}$$\textrm{b a s e} = \sqrt[\textrm{p o w e r}]{\textrm{r e s \underline{t}}}$

This is called "root".
The same in another form is
$\text{base}={\left(\text{exp. result}\right)}^{\frac{1}{\text{power}}}$$\textrm{b a s e} = {\left(\textrm{\exp . r e s \underline{t}}\right)}^{\frac{1}{\textrm{p o w e r}}}$
This is exponent to a fraction.

If exponentiation-result and base are given, then
$\text{power}={\mathrm{log}}_{\text{base}}\left(\text{exp.result}\right)$$\textrm{p o w e r} = {\log}_{\textrm{b a s e}} \left(\textrm{\exp . r e s \underline{t}}\right)$
This inverse is called "logarithm".

The word "root" means: basic source or origin of something.

Roots : Root of a number to a given power of root is the base of the exponent with the given power.

$\sqrt{8}=2$$\sqrt{8} = 2$
$3$$3$ is the power of root
$8$$8$ is the number for which root is calculated
$2$$2$ is the result of root

$\sqrt{8}=2$$\sqrt{8} = 2$ implies that ${2}^{3}=8$${2}^{3} = 8$, and the operation root finds the base in the equivalent exponent.

evaluating

Finding Root (First Principles) : Root of a number is the base in the equivalent exponent.

eg: $\sqrt{64}$$\sqrt{64}$ is seen as the exponent $64={4}^{3}$$64 = {4}^{3}$. The base is $4$$4$ and so $\sqrt{64}=4$$\sqrt{64} = 4$

It is noted that ${81}^{\frac{1}{4}}$${81}^{\frac{1}{4}}$ is "$\sqrt{81}$$\sqrt{81}$".

Find $\sqrt{125}$$\sqrt{125}$
The answer is "$5$$5$".

To find $\sqrt{125}$$\sqrt{125}$, perform prime-factorization on the given value.
$125=5×5×5$$125 = 5 \times 5 \times 5$
From this, it is evident that $125={5}^{3}$$125 = {5}^{3}$. By first principles, $\sqrt{125}=5$$\sqrt{125} = 5$

Find $\sqrt{36}$$\sqrt{36}$
The answer is "$6$$6$".

To find $\sqrt{36}$$\sqrt{36}$, perform prime-factorization on the given value.

$36=2×2×3×3$$36 = 2 \times 2 \times 3 \times 3$
re-arrage such that the factors are grouped
$36=\left(2×3\right)×\left(2×3\right)$$36 = \left(2 \times 3\right) \times \left(2 \times 3\right)$
There are two groups equal to the power of the root $2$$2$.
pick one group and compute the result.
By first principles, $\sqrt{36}=2×3=6$$\sqrt{36} = 2 \times 3 = 6$

Finding Roots (Simplified Procedure) : To find roots of a number, express the number in prime factors and group the factors.

eg: $\sqrt{1000}$$\sqrt{1000}$ $=\sqrt{2×2×2×5×5×5}$$= \sqrt{2 \times 2 \times 2 \times 5 \times 5 \times 5}$ $=10$$= 10$

summary

Roots : Root of a number to a given power of root is the base of the exponent with the given power.

$\sqrt{8}=2$$\sqrt{8} = 2$
$3$$3$ is the power of root
$8$$8$ is the number for which root is calculated
$2$$2$ is the result of root

$\sqrt{8}=2$$\sqrt{8} = 2$ implies that ${2}^{3}=8$${2}^{3} = 8$, and the operation root finds the base in the equivalent exponent.

Finding Root (First Principles) : Root of a number is the base in the equivalent exponent.

eg: $\sqrt{64}$$\sqrt{64}$ is seen as the exponent $64={4}^{3}$$64 = {4}^{3}$. The base is $4$$4$ and so $\sqrt{64}=4$$\sqrt{64} = 4$

Finding Roots (Simplified Procedure) : To find roots of a number, express the number in prime factors and group the factors.

eg: $\sqrt{1000}$$\sqrt{1000}$ $=\sqrt{2×2×2×5×5×5}$$= \sqrt{2 \times 2 \times 2 \times 5 \times 5 \times 5}$ $=10$$= 10$

Outline

The outline of material to learn "Exponents" is as follows. Note: click here for detailed outline of Exponents s

→   Representation of Exponents

→   Inverse of exponent : root

→   Inverse of exponent : Logarithm

→   Common and Natural Logarithms

→   Exponents Arithmetics

→   Logarithm Arithmetics

→   Formulas

→   Numerical Expressions

→   PEMA / BOMA

→   Squares and Square roots

→   Cubes and Cube roots