Questions about Exponentiation and roots and logarithms.

Question

in this page a few questions I want to ask you about the Exponentiation and roots and logarithms:

What and how the Exponentiation definition can be defined by real numbers.?

What is the overall definition roots?(why ?$ \sqrt[n]{ x} = x^{ \frac{1}{n} } $)

What is the general definition of logarithm?

In the notation ($ \sqrt[n]{x} $), why must $n$ be a natural number greater than 2?

Answer 1

$\log_ab=c$ is defined to mean $a^c=b$.

$\root n\of x=x^{1/n}$ because we want it to be true that $(x^a)^b=x^{ab}$.

Answer 2

We will define the symbol $a^b$ for $a>0$ and any real $b$.

• Natural exponent First let $b$ be a natural number. In this case we define $a^b = a \cdot a \cdot a \cdots a$ for $b$ occurrences of $a$. Also $a^1 = a$ and $a^0 = 1$. Note the properties: (i) $a^{b_1}\cdot a^{b_2} = a ^{b_1 + b_2}$ and (ii) $(a^{b_1})^{b_2}= a^{b_1 \cdot b_2}$.
• Positive rational exponents Now let $b$ be rational. Let's assume $b = \frac{p}{q}$, where $p$ and $q$ are positive integers in lowest terms. What should $a^{p/q}$ mean? If we require that (ii) holds for rational exponents we get $$\left(a^\frac{p}{q}\right)^q = a^{\frac{p}{q}\cdot q} = a^p$$ So $a^{p/q}$ is the unique positive solution of the equation $x^q=a^p, x>0$ (1). This number is called the q-th root of $a^p$ and is denoted $\sqrt[q]{a^p}$. Therefore $a^{p/q} = \sqrt[q]{a^p}$. This number is well defined since for $x>0$ the function $y=x^p$ is well defined according to the definition for natural exponents. Moreover it is 1-1 and its image is $(0,+\infty)$ meaning exactly that equation (1) has exactly one real positive solution for each possible $a^p$. (Instead of writing $\sqrt[2]{x}$ we write $\sqrt{x}$, and instead of $\sqrt[1]{x}$ we write $x$.)
• Real positive exponents Lets generalise to the positive reals i.e. let b be any real positive number. There exists a sequence $\left\lbrace b_n\right\rbrace_{n\in\mathbb{N}}$ of rationals such that it converges to b as n get large, i.e. $\lim_{n\rightarrow \infty} b_n = b$. (One such sequence is the decimal expansion of $b$ to $n$ digits. For example consider the sequence of rationals: 3, 3.1, 3.14, 3.141, $\cdots$. This sequence converges to $\pi$, but each individual value is rational). Define the new sequence $r_n = a^{b_n}$, for all $n\in\mathbb{N}$. This exists due to the definition for rational exponents. It is provable that this sequence converges to a real number $r$ as n gets large. This number is the definition of $a^b$. I.e. $a^b = \lim_{n\rightarrow \infty} a ^{b_n}$ where $\left\lbrace b_n\right\rbrace_{n\in\mathbb{N}}$ is any random sequence of rationals that converges to $b$.
• Negative exponents We simply define that $a^{-b} = \frac{1}{a^b}$ for any $a,b>0$.
• Last we shall treat the case where $a=0$. If $b>0$, then $a^b = 0^b = 0$, and if $b\leq 0$ then $a^b = 0 ^b$ is left undefined.
• For negative bases $a<0$ we only define integer exponents (b is integer) according to the natural exponent case and by requiring that $a^{-b} = \frac{1}{a^b}$ holds.

Now about logarithms, as stated we define $\log_a b = c$ if and only if $a^c =b$. To understand the relationship between exponentiation, logarithms and roots consider the equality $2^3 = 8$ which shows a relation between 2, 3, and 8:

• To get 8 out of 2 and 3 exponentiate: $8=2^3$
• To get 2 out of 8 and 3 take the root: $2 = \sqrt[3]{8}$
• To get 3 out of 8 and 2 take the logarithm: $3 = \log_2 8$

Sorry for the length. Hope I helped.