Properties Of Exponents

I. Zero Exponent

How much is some number $b$ raised to the power $0$, that is, $b^0$

We saw that for any number $b, m$ and $n$

$b^m \div b^n = b^{(m-n)}$.

Let us find out. How much is $b^5 \div b^5$.

Expanding this as normal power we get:

$b^5 \div b^5$

$= \dfrac{b \times b \times b \times b \times b}{b \times b \times b \times b \times b}$

As you see, there are 5 $b'$s in the numerator and 5 $b'$s in the denominator, so all of them will cancel out:

$= \dfrac{\cancel{b} \times \cancel{b} \times \cancel{b} \times \cancel{b} \times \cancel{b}}{\cancel{b} \times \cancel{b} \times \cancel{b} \times \cancel{b} \times \cancel{b}}$

leaving us with:

$\dfrac{1}{1} = 1$

Now, let us go back to our first statement, that $b^m \div b^n = b^{(m-n)}$.

In our example both $m$ and $n$ are 5.

So, we get $b^5 \div b^5 = b^{(5-5)} = b^0$

So, by expanding we saw that $b^5 \div b^5 = 1$ and from our first fact we see that $b^5 \div b^5 = b^0$.

So, we can say that $b^0 = 1$.

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We have seen this for a specific value of $5$ for easy understanding. But it is true for any general value of $m$ and $n$.

If we put $m = n$ then our first equation, we get:

$b^m \div b^m = b^{(m-m)} = b^0$,

We also know that

$b^m \div b^m = \dfrac{b^m}{b^m} = \dfrac{\cancel{b^m}}{\cancel{b^m}} = 1$

$\texttip{\therefore}{therefore} b^0 = 1$.

II. Negative Exponents

When we write a number as $b^{-n}$ where $n$ is a positive integer, what does it mean?

Let us try to understand this better with an example. How much is $2^{-3}$?

We know that we can write $-3$ as $0 - 3$, therefore,

$2^{-3} = 2^{0-3}$

Since, we know the relationship, $b^m \div b^n = b^{(m-n)}$, we can write:

$2^{0-3} = \dfrac{2^0}{2^3}$

Again, we know that $2^0 = 1$, therefore

$\dfrac{2^0}{2^3} = \dfrac{1}{2^3}$

Therefore, we finally get:
$2^{-3} = \dfrac{1}{2^3}$

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Now, let us go back to the general case of $b^{-n}$.

$b^{-n}$

$= b^{0 - n}$

$= \dfrac{b^0}{b^n}$

$= \dfrac{1}{b^n}$

III. Fraction Exponents

When we have a fraction as an exponent, like $b^{\frac{1}{n}}$, what does it mean? Like before, let us start with an example.

Let say, how much is $16^{\frac{1}{4}}$?

Let $16^{\frac{1}{4}} = x$

Let us multiply four $x$s together and see what we get:
$x \times x \times x \times x = 16^{\frac{1}{4}} \times 16^{\frac{1}{4}} \times 16^{\frac{1}{4}} \times 16^{\frac{1}{4}}$

$\Rightarrow x^4 = 16^{\left(\frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} \right)}$

$\Rightarrow x^4 = 16^{\left( 1 \right)} = 16$

We know that the inverse operation of exponentiation is the root function.

$\therefore x = \sqrt[4]{16}$

Since we started by taking $16^{\frac{1}{4}} = x$:

$16^{\frac{1}{4}} = x = \sqrt[4]{16}$

$\Rightarrow 16^{\frac{1}{4}} = \sqrt[4]{16}$

In general we can write that $b^{\frac{m}{n}} = \left( \sqrt[n]{b} \right)^m = \sqrt[n]{b^m}$

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Now, let us solve the following problem before we move to the next section:

--------- Reference to question: 70410412-3f6b-479f-8989-91e3ed32fc0f ---------

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IV. Equality Of Exponents

We will briefly look into another property of bases and exponents here.

If $a^x = a^y$, where $a \ne 0, \pm 1$ it is always true that $x = y$.

Now let us see what happens when there are multiple bases in our expressions.

If an integer $N$ has factors (prime or composite) $a$, $b$, $c$ where $a$, $b$ and $c$ are mutually co-prime, then, there is only one way of writing $N = a^xb^yc^z$. In other words all of $x$, $y$ and $z$ will have single solution.

We can extend this to the following concept:

For mutually coprime numbers $a$, $b$ and $c$

if $a^xb^yc^z = a^pb^qc^r$ then $x = p$, $y = q$ and $z = r$

For example if it is given that $2^x3^y5^z = 2^73^25^9$ then we can say that $x = 7$, $y = 2$ and $z = 9$

However, if $a$, $b$ or $c$ are not mutually coprime, and it is possible to rearrange the prime factors of one or the bases into the other bases, then this is not true.

Let us see the following example:

Given that $2^x6^y9^z = 2^76^49^2$ can we say that $x = 7$, $y = 4$ and $z = 7$?

$We\ cannot!!!$

We can rewrite $2^76^49^2$ as $2^96^29^3$

Therefore our original equation can become:

$2^x6^y9^z = 2^76^49^2 = 2^96^29^3$

Likewise there can be many more ways we can rewrite the right hand side expression using bases of $2$, $6$ and $9$.

Therefore, we cannot say what would be the value of $x$, $y$ and $z$.