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why does 3 to the power of 0 always 1

Guest May 25, 2017

Best Answer 

 #1
avatar+1493 
+2

As a generalization, \(x^0=1\hspace{1cm},x\neq0\)

 

This table may help you understand why 3^0=1

\(3^{10}\)   3*3*3*3*3*3*3*3*3*3 \(59049\)
\(3^9\)   3*3*3*3*3*3*3*3*3 \(19683\)
\(3^8\)   3*3*3*3*3*3*3*3 \(6561\)
\(3^7\)   3*3*3*3*3*3*3 \(2187\)
\(3^6\)   3*3*3*3*3*3 \(729\)
\(3^5\)   3*3*3*3*3 \(243\)
\(3^4\)   3*3*3*3 \(81\)
\(3^3\)   3*3*3 \(27\)
\(3^2\)   3*3 \(9\)
\(3^1\)   3 \(3\)
\(3^0\)   ? ?

 

Do you notice a pattern? I do. As you go down the list, you can divide by three to get to the next value. Therefore, if 3^1=3, all you have to do to get the next value is to divide by three. 3^1/3=1, so 3^0=1.

 

Here's another way of thinking about it. This method works for any number to the power of 0:

 

\(1=\frac{x^n}{x^n}=x^{n-n}=x^0\hspace{1cm},x\neq0\)

TheXSquaredFactor  May 25, 2017
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1+0 Answers

 #1
avatar+1493 
+2
Best Answer

As a generalization, \(x^0=1\hspace{1cm},x\neq0\)

 

This table may help you understand why 3^0=1

\(3^{10}\)   3*3*3*3*3*3*3*3*3*3 \(59049\)
\(3^9\)   3*3*3*3*3*3*3*3*3 \(19683\)
\(3^8\)   3*3*3*3*3*3*3*3 \(6561\)
\(3^7\)   3*3*3*3*3*3*3 \(2187\)
\(3^6\)   3*3*3*3*3*3 \(729\)
\(3^5\)   3*3*3*3*3 \(243\)
\(3^4\)   3*3*3*3 \(81\)
\(3^3\)   3*3*3 \(27\)
\(3^2\)   3*3 \(9\)
\(3^1\)   3 \(3\)
\(3^0\)   ? ?

 

Do you notice a pattern? I do. As you go down the list, you can divide by three to get to the next value. Therefore, if 3^1=3, all you have to do to get the next value is to divide by three. 3^1/3=1, so 3^0=1.

 

Here's another way of thinking about it. This method works for any number to the power of 0:

 

\(1=\frac{x^n}{x^n}=x^{n-n}=x^0\hspace{1cm},x\neq0\)

TheXSquaredFactor  May 25, 2017

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