Definition: $a^x$ is defined as:
1) For $x\;\in\mathbb{Z^+}$:
$\qquad\quad a^x = \underbrace{(a.a.a.....a)}_\text{$x \text{ times}$}\\[6pt]
\qquad\quad a^{-x} = \dfrac{1}{a^x}$
$\;\;\;$For $a>0,\\[6pt]
\text{If}\;\;\quad\quad y^{x} = a\\
\text{Then,} \quad a^{1/x} = y$
2) For $x\;\in\mathbb{Q}$, then $x = p/q$, where, $p, q\;\in\mathbb{Z^+}$
$\;\;\;$For $a>0,\\[6pt]
\text{If}\;\;\quad\quad\;\; a^p = b^q\\[3pt]
\text{Then,} \quad a^{\tfrac{p}{q}} = b$
3) For $x\;\in\mathbb{R}$ and $a>0$:
$\;\;\; a^x$ for irrational numbers is defined in a way so as to have the continuous exponential function.
Showing posts with label Exponentiation. Show all posts
Showing posts with label Exponentiation. Show all posts
Power Zero
Theorem: $\; a^0 = 1\;\;\forall\; a\neq 0$
Prerequisites:
Sum of the powers property (proof)
Proof:
$\qquad\quad \begin{align}a^0 &= a^{n-n} \qquad\quad\;\;\;\text{ where $n\in\mathbb{Z^+}$}\\
&= a^n.a^{-n}\qquad\quad\text{(by sum of the power property)}\\
&= \dfrac{a^n}{a^n} = 1\end{align}$
Hence the result
Recommended:
Sum of the powers
Product of the powers
Power of the product
Prerequisites:
Sum of the powers property (proof)
Proof:
$\qquad\quad \begin{align}a^0 &= a^{n-n} \qquad\quad\;\;\;\text{ where $n\in\mathbb{Z^+}$}\\
&= a^n.a^{-n}\qquad\quad\text{(by sum of the power property)}\\
&= \dfrac{a^n}{a^n} = 1\end{align}$
Hence the result
Recommended:
Sum of the powers
Product of the powers
Power of the product
Power Of The Product
Theorem: $\;(ab)^x = a^x.b^x\;\; \forall\; a,b > 0 \;\text{ and }\; x\in\mathbb{R}$
Prerequisites:
Log of product property (proof)
Log of powers property (proof)
Proof:
Taking log of $(ab)^x$:
$\qquad\quad\begin{align} \log{(ab)^x} &= x\log{(ab)}\qquad\;\;\qquad\qquad\qquad\text{(by log of power property)}\\
&= x(\log{a} + \log{b})\qquad\qquad\qquad\text{(by log of product property)}\\
&= x\log{a} + x\log{b}\\
&= \log{a^x} + \log{b^x}\qquad\;\qquad\qquad\text{(by log of power property)}\\
&= \log{a^x.b^x}\qquad\qquad\;\;\qquad\qquad\text{(by log of product property)}\end{align}$
Taking antilog on both sides:
$\qquad\quad(ab)^x = a^x.b^x$
Hence the result
Recommended:
Sum of the powers
Product of the powers
Log of product
Log of powers
Prerequisites:
Log of product property (proof)
Log of powers property (proof)
Proof:
Taking log of $(ab)^x$:
$\qquad\quad\begin{align} \log{(ab)^x} &= x\log{(ab)}\qquad\;\;\qquad\qquad\qquad\text{(by log of power property)}\\
&= x(\log{a} + \log{b})\qquad\qquad\qquad\text{(by log of product property)}\\
&= x\log{a} + x\log{b}\\
&= \log{a^x} + \log{b^x}\qquad\;\qquad\qquad\text{(by log of power property)}\\
&= \log{a^x.b^x}\qquad\qquad\;\;\qquad\qquad\text{(by log of product property)}\end{align}$
Taking antilog on both sides:
$\qquad\quad(ab)^x = a^x.b^x$
Hence the result
Recommended:
Sum of the powers
Product of the powers
Log of product
Log of powers
Product Of The Powers
Theorem: $\;\;\big(a^{x}\big)^y = a^{\displaystyle (xy)}\quad\forall\; x, y \in \mathbb{R} \text{ and } a > 0$
Prerequisites:
Exponentiation (definition)
Sum of the Powers Property (proof)
Continuity of exponential function
Proof:
Case I: When $y\in\mathbb{Z}$
$\qquad\quad\begin{align}\big(a^{x}\big)^y &= a^x.a^x.a^x....a^x\qquad\;\;\text{($y$ times)}\\
&= a^{\displaystyle (x+x+x...x)}\qquad\text{(by sum of powers property)}\\
&= a^{\displaystyle (xy)}\end{align}$
Case II: When $y\in\mathbb{Q}$
Let $y=p/q$, where $p, q\in\mathbb{Z}$
Let $\quad a^{\tfrac{p}{q}x} = b\qquad\qquad\qquad\cdots\text{(1)}$
$\therefore\quad\;\; \big(a^{\tfrac{x}{q}}\big)^p = b\qquad\qquad\qquad\text{(from case I)}\\
\Rightarrow\quad\;\; a^{\tfrac{x}{q}} = b^{\tfrac{1}{p}}\qquad\qquad\qquad\;\text{(by definition of exponentiation)}$
Raising both sides to the power $q$,
$\therefore\quad\;\; \big(a^{\tfrac{x}{q}}\big)^q = \big(b^{\tfrac{1}{p}}\big)^q\\
\Rightarrow\quad\;\; a^{\tfrac{x}{q}q} = b^{\tfrac{q}{p}}\qquad\qquad\qquad\text{(from case I)}\\
\Rightarrow\quad\;\; a^{x} = b^{\tfrac{q}{p}}$
Let $q/p=r$
$\therefore\quad\quad a^{x} = b^r\\
\Rightarrow\quad\;\; \big(a^x\big)^{\tfrac{1}{r}} = b\qquad\qquad\qquad\text{(by definition of exponentiation)}\\
\Rightarrow\quad\;\; \big(a^x\big)^{\tfrac{1}{r}} = b = a^{\tfrac{p}{q}x}\qquad\quad\!\!\text{(from $(1)$)}$
Case III: When $y\in\mathbb{R}$
If $y$ is irrational, then in any neighbourhood of $y$, there exists infinitely many rational numbers. Since at these rational numbers, the given equality holds, hence by the continuity of the exponential functions, the equality must also hold at $y$.
Therefore the result holds at all real values of $x$ and $y$.
Recommended:
Sum of the powers
Power of the product
Log of product
Log of powers
Prerequisites:
Exponentiation (definition)
Sum of the Powers Property (proof)
Continuity of exponential function
Proof:
Case I: When $y\in\mathbb{Z}$
$\qquad\quad\begin{align}\big(a^{x}\big)^y &= a^x.a^x.a^x....a^x\qquad\;\;\text{($y$ times)}\\
&= a^{\displaystyle (x+x+x...x)}\qquad\text{(by sum of powers property)}\\
&= a^{\displaystyle (xy)}\end{align}$
Case II: When $y\in\mathbb{Q}$
Let $y=p/q$, where $p, q\in\mathbb{Z}$
Let $\quad a^{\tfrac{p}{q}x} = b\qquad\qquad\qquad\cdots\text{(1)}$
$\therefore\quad\;\; \big(a^{\tfrac{x}{q}}\big)^p = b\qquad\qquad\qquad\text{(from case I)}\\
\Rightarrow\quad\;\; a^{\tfrac{x}{q}} = b^{\tfrac{1}{p}}\qquad\qquad\qquad\;\text{(by definition of exponentiation)}$
Raising both sides to the power $q$,
$\therefore\quad\;\; \big(a^{\tfrac{x}{q}}\big)^q = \big(b^{\tfrac{1}{p}}\big)^q\\
\Rightarrow\quad\;\; a^{\tfrac{x}{q}q} = b^{\tfrac{q}{p}}\qquad\qquad\qquad\text{(from case I)}\\
\Rightarrow\quad\;\; a^{x} = b^{\tfrac{q}{p}}$
Let $q/p=r$
$\therefore\quad\quad a^{x} = b^r\\
\Rightarrow\quad\;\; \big(a^x\big)^{\tfrac{1}{r}} = b\qquad\qquad\qquad\text{(by definition of exponentiation)}\\
\Rightarrow\quad\;\; \big(a^x\big)^{\tfrac{1}{r}} = b = a^{\tfrac{p}{q}x}\qquad\quad\!\!\text{(from $(1)$)}$
Case III: When $y\in\mathbb{R}$
If $y$ is irrational, then in any neighbourhood of $y$, there exists infinitely many rational numbers. Since at these rational numbers, the given equality holds, hence by the continuity of the exponential functions, the equality must also hold at $y$.
Therefore the result holds at all real values of $x$ and $y$.
Recommended:
Sum of the powers
Power of the product
Log of product
Log of powers
Sum Of The Powers
Theorem: $\;\;a^{x+y} = a^x.a^y \quad\forall\; x, y \in \mathbb{R} \text{ and } a > 0$
Prerequisites:
Exponentiation (definition)
Continuity of exponential function
Proof:
Case I: when $x, y \in\mathbb{Z^+}$
$\qquad\quad \begin{align} a^{x+y} &= a.a.a.a......a\quad \text{($x+y$ times)}\\
&= \underbrace{(a.a.a.....a)}_\text{$x \text{ times}$}.\underbrace{(a.a.a.....a)}_\text{$y\text{ times}$}\\
&= a^x.a^y\end{align}$
Case II: when $x, y \in\mathbb{Z^-}$
Let $x_1 = -x, \;\; y_1 = -y$ and $1/a = b$
$\qquad\quad \begin{align} a^{x+y} &= a^{-(x_1+y_1)}\\\\
&= \left(\dfrac{1}{a}\right)^{(x_1+y_1)}\\\\
&= b^{(x_1+y_1)}\end{align}$
Hence the case is reduced to the previous case and can be proved in a similar manner.
Case III: when $x \in\mathbb{Z^+}$ and $y \in\mathbb{Z^-}$
Let $y_1 = -y$
$\qquad\quad \begin{align} a^{x+y} &= a^{x-y_1}\\
&= a.a.a....a\quad\text{($x-y_1$ times)}\\\
&= \dfrac{a.a.a....a}{a.a.a....a}\begin{matrix}\text{($x$ times)}\\
\text{($y_1$ times)}\end{matrix}\\\
&= \dfrac{a^x}{a^{y_1}}\\\
&= a^x.a^{-y_1}\\
&= a^x.a^y\end{align}$
Case IV: When atleast one of $x, y$ equal to $0$
Let $y = 0$
$\qquad\quad \begin{align}a^0 &= a^{n-n} \qquad\quad\;\;\;\text{ where $n\in\mathbb{Z^+}$}\\
&= a^n.a^{-n}\qquad\quad\text{(from case III)}\\
&= \dfrac{a^n}{a^n} = 1\end{align}$
$\begin{align}\text{Now, }\quad a^{x+y} &= a^{x+0} = a^x\\
&= a^x.1 = a^x.a^0\\
&= a^x.a^y\end{align}$
Case V: When $x, y \in\mathbb{Q}$, i.e., $x, y$ are rational numbers
As a corollary of case I, for $c \in\mathbb{Z^+}$
$\qquad\quad\begin{align}\big(a^{b}\big)^c &= a^b.a^b.a^b....a^b\qquad\text{($c$ times)}\\
&= a^{(b+b+b...b)}\\
&= a^{\displaystyle (bc)}\end{align}$
Let $x = p_1/q_1$ and $y = p_2/q_2$ where $p_1, p_2, q_1, q_2 \in\mathbb{Z}\;\text{ and }\; q_1, q_2\neq 0$
Let $q' = q_1q_2$,
$ \begin{align} \text{Now, }\quad a^{x+y} &= a^{\tfrac{p_1}{q_1}+ \tfrac{p_2}{q_2}}\\\\
&= a^{\tfrac{p_1q_2 + p_2q_1}{q'}}\\\\
&= \Big( a^{(1/q')}\Big) ^{(p_1q_2 + p_2q_1)}\qquad\qquad\qquad\text{(from the above corollary)}\\\\
&= \Big( a^{(1/q')}\Big) ^{(p_1q_2)}\Big( a^{(1/q')}\Big) ^{(p_2q_1)}\qquad\text{(from case I)}\\\\
&= a^{\left( \tfrac{p_1q_2}{q'}\right)}a^{\left( \tfrac{p_2q_1}{q'}\right)}\\\\
&= a^{\tfrac{p_1}{q_1}}.a^{\tfrac{p_2}{q_2}}\\\\
&= a^x.a^y\end{align}$
Case VI: When $x, y\in\mathbb{R}$
If $x$ and $y$ are irrational, then in any neighbourhood of $x$ and $y$, there exists infinitely many rational numbers. Since at these rational numbers, the given equality holds, hence by the continuity of the exponential functions, the equality must also hold at $x$ and $y$.
Therefore the result holds at all real values of $x$ and $y$.
Recommended:
Product of the powers
Power of the product
Log of product
Log of powers
Prerequisites:
Exponentiation (definition)
Continuity of exponential function
Proof:
Case I: when $x, y \in\mathbb{Z^+}$
$\qquad\quad \begin{align} a^{x+y} &= a.a.a.a......a\quad \text{($x+y$ times)}\\
&= \underbrace{(a.a.a.....a)}_\text{$x \text{ times}$}.\underbrace{(a.a.a.....a)}_\text{$y\text{ times}$}\\
&= a^x.a^y\end{align}$
Case II: when $x, y \in\mathbb{Z^-}$
Let $x_1 = -x, \;\; y_1 = -y$ and $1/a = b$
$\qquad\quad \begin{align} a^{x+y} &= a^{-(x_1+y_1)}\\\\
&= \left(\dfrac{1}{a}\right)^{(x_1+y_1)}\\\\
&= b^{(x_1+y_1)}\end{align}$
Hence the case is reduced to the previous case and can be proved in a similar manner.
Case III: when $x \in\mathbb{Z^+}$ and $y \in\mathbb{Z^-}$
Let $y_1 = -y$
$\qquad\quad \begin{align} a^{x+y} &= a^{x-y_1}\\
&= a.a.a....a\quad\text{($x-y_1$ times)}\\\
&= \dfrac{a.a.a....a}{a.a.a....a}\begin{matrix}\text{($x$ times)}\\
\text{($y_1$ times)}\end{matrix}\\\
&= \dfrac{a^x}{a^{y_1}}\\\
&= a^x.a^{-y_1}\\
&= a^x.a^y\end{align}$
Case IV: When atleast one of $x, y$ equal to $0$
Let $y = 0$
$\qquad\quad \begin{align}a^0 &= a^{n-n} \qquad\quad\;\;\;\text{ where $n\in\mathbb{Z^+}$}\\
&= a^n.a^{-n}\qquad\quad\text{(from case III)}\\
&= \dfrac{a^n}{a^n} = 1\end{align}$
&= a^x.1 = a^x.a^0\\
&= a^x.a^y\end{align}$
Case V: When $x, y \in\mathbb{Q}$, i.e., $x, y$ are rational numbers
As a corollary of case I, for $c \in\mathbb{Z^+}$
$\qquad\quad\begin{align}\big(a^{b}\big)^c &= a^b.a^b.a^b....a^b\qquad\text{($c$ times)}\\
&= a^{(b+b+b...b)}\\
&= a^{\displaystyle (bc)}\end{align}$
Let $x = p_1/q_1$ and $y = p_2/q_2$ where $p_1, p_2, q_1, q_2 \in\mathbb{Z}\;\text{ and }\; q_1, q_2\neq 0$
Let $q' = q_1q_2$,
$ \begin{align} \text{Now, }\quad a^{x+y} &= a^{\tfrac{p_1}{q_1}+ \tfrac{p_2}{q_2}}\\\\
&= a^{\tfrac{p_1q_2 + p_2q_1}{q'}}\\\\
&= \Big( a^{(1/q')}\Big) ^{(p_1q_2 + p_2q_1)}\qquad\qquad\qquad\text{(from the above corollary)}\\\\
&= \Big( a^{(1/q')}\Big) ^{(p_1q_2)}\Big( a^{(1/q')}\Big) ^{(p_2q_1)}\qquad\text{(from case I)}\\\\
&= a^{\left( \tfrac{p_1q_2}{q'}\right)}a^{\left( \tfrac{p_2q_1}{q'}\right)}\\\\
&= a^{\tfrac{p_1}{q_1}}.a^{\tfrac{p_2}{q_2}}\\\\
&= a^x.a^y\end{align}$
Case VI: When $x, y\in\mathbb{R}$
If $x$ and $y$ are irrational, then in any neighbourhood of $x$ and $y$, there exists infinitely many rational numbers. Since at these rational numbers, the given equality holds, hence by the continuity of the exponential functions, the equality must also hold at $x$ and $y$.
Therefore the result holds at all real values of $x$ and $y$.
Recommended:
Product of the powers
Power of the product
Log of product
Log of powers
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