Let’s break this down. Consider the example below with a base of 7. It can help to think of this as: 7

Question

3
times what value equals itself? Keep in mind that any number multiplied by 1 equals itself. What value of 7
0
would make the equation below true?
7
3
· 7
0
=7
3

7
3
·
=7
3


a
0
= ?

Answer 1

To solve this problem, we are trying to find the value of \( a \) such that the equation \( 7^3 \cdot 7^0 = 7^3 \) holds true.

We can start by using the properties of exponents. Recall the following property:

\[ a^m \cdot a^n = a^{m+n} \]

In this case, we have:

\[ 7^3 \cdot 7^0 = 7^{3+0} = 7^3 \]

Since the left side simplifies to \( 7^3 \), we see that the equation holds true regardless of the value of \( 7^0 \) as long as \( 7^0 = 1 \).

Therefore, to answer your question, the value of \( a \) in \( 7^0 = a \) is:

\[ a = 1 \]

In conclusion, \( 7^0 = 1 \).