# What is 5^0?

Dec 3, 2014

As Samiha explained, any number raised to the power of 0 is equal to 1. I am going to show how that works out.

By the laws of exponents, when the bases are equal, the powers can be added up for multiplication and subtracted for division.

i.e.,

${x}^{a} \cdot {x}^{b} = {x}^{a + b}$

${x}^{a} / {x}^{b} = {x}^{a - b}$

As an example,

${2}^{1} \cdot {2}^{4} = {2}^{1 + 4} = {2}^{5}$

and ${2}^{1} / {2}^{4} = {2}^{1 - 4} = {2}^{-} 3$

I'll be using the second property.

Now, we know that any number divided by itself is equal to 1. Just as an example,

$1 = {3}^{2} / {3}^{2}$

But, applying the second property,

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

Thus, it can be concluded that ${3}^{0} = 1$. In fact, this would hold true for any number $x$.

$1 = {x}^{n} / {x}^{n} = {x}^{n - n} = {x}^{0}$

Thus, ${x}^{0} = 1$ for any number $x$.

I am going to show the same in another form.

Consider the following numbers arranged in a sequence (I have written their equivalents below).

${5}^{1} , {5}^{2} , {5}^{3} , {5}^{4} , \ldots$
$5 , 25 , 125 , 625 , \ldots$

It can be seen that the next term of the sequence can be obtained by multiplying the last one by 5.
Another way of putting this is that the previous term of a sequence can be obtained by dividing by 5.

The logical precedent of ${5}^{1}$ in the first sequence would be ${5}^{0}$.
Similarly, the logical precedent of $5$ in the second sequence would be $\frac{5}{5} = 1$.

Since they both are the same sequence, it can be concluded that

${5}^{0} = 1$

This would again hold true for any number $x$.

So, ${x}^{0} = 1$ for any number $x$.