Please explain why $10$ raised to the power of $0$ ($10^0$) is equal to $1$

Question

Please explain why $10$ raised to the power of $0$ i.e., $10^0$ is equal to $1$ and $10^1$ is simply equal to $10$, then the numbers between $1$ and $10$ can be written as $10$ raised to some power between $0$ and $1$.

In the same way that the sum of no numbers is zero, the product of no numbers is one. Your second result doesn't "follow from" the first without extending the definition of exponentiation. — Nick, Jan 08 '14 at 14:17

Lutz Lehmann · Answer 1 · 2014-01-08T14:27:58.637

1

This is a convention that arises most naturally from the fact that polynomials are linear combinations over a geometric sequence $1,x,x^2,x^3,x^4,x^5,...$. So it comes quite natural to write the first two terms as $x^0$ and $x^1$. This is then also compatible with the power laws $x^{m+n}=x^m\cdot x^n$ for all integers $m,n$, including the negative ones.

edited Jan 08 '14 at 14:27

answered Jan 08 '14 at 14:18

Lutz Lehmann

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score 1 · Answer 2 · answered Jan 08 '14 at 14:38

Just as examples like

$$a^2a^3=(a\cdot a)(a\cdot a\cdot a)=a\cdot a\cdot a\cdot a\cdot a=a^5$$

lead to the general rule

$$a^ma^n=a^{m+n}$$

for positive integers $m$ and $n$, so too do examples like

$${a^5\over a^2}={a\cdot a\cdot a\cdot a\cdot a\over a\cdot a}=a\cdot a\cdot a=a^3$$

(with $a\not=0$) lead to the general rule

$${a^n\over a^m} = a^{n-m}$$

for positive integers with $n\gt m$. The conventions $a^0=1$ and $a^{-n}=1/a^n$ (again assuming $a\not=0$) are what you get if you want to make the general rules apply to all integers $m$ and $n$, without restriction.

score 0 · Answer 3 · answered Jan 08 '14 at 14:21

There are many ways to explain this.One way is :

$$\frac{a^n}{a^n} = \frac{a^n}{a^n}$$ $$a^{(n-n)} = 1$$ $$a^0 = 1$$ where $a$ not equal to $0$.

Since logarithmic function is a strictly increasing function, you can conclude that if: $$10^0 \leq 10^x \leq 10^1$$ then, $$0 \leq x \leq 1$$

score 0 · Answer 4 · answered Jan 08 '14 at 14:22

We know that when we multiply powers with same base, then we add the exponents and when we divide we subtract them so we have:

$$1 = \frac{a^x}{a^x} = a^{x-x} = a^0$$

Also we know that the exponential function is strictly increasing for every base bigger than 1. Since we now that $10^0 = 1$ and $10^1 = 10$, then every number between $1$ and $10$ can be writen as $10^x$, where $x \in (0,1)$

score 0 · Answer 5 · answered Jan 08 '14 at 15:14

The other answers all illuminate various reasons for the equality, but not knowing how much algebra you want to put into it, here's what we were taught in elementary school (which does not really add anything that the other answers haven't): increasing the power of $10$ by one is the same as multiplying by $10$. For instance, $10^4 = 10 \cdot 10^3$. On the other hand, decreasing the power by one is the same as dividing by $10$, e.g. $10^3 = 10^4 / 10$. Extending this logic, we should have that $10^1 = 10^2 / 10 = 100/10 = 10$, and $10^0 = 10^1 / 10 = 10 / 10 = 1$.

Please explain why $10$ raised to the power of $0$ ($10^0$) is equal to $1$

5 Answers5