Question

With the zero power rule of exponents

Is 1/2(9 - 7^0) + (-29)^0 equivalent to 1/2(9 - 1) - 1?

Answer 1

Let's evaluate both expressions step by step using the zero power rule of exponents, which states that any non-zero number raised to the power of zero is equal to 1.

**Starting with the left side:**

\[
\frac{1}{2}(9 - 7^0) + (-29)^0
\]

1. Calculate \(7^0\):
\[
7^0 = 1
\]

2. Substitute this value back into the expression:
\[
\frac{1}{2}(9 - 1) + (-29)^0
\]

3. Now calculate \((-29)^0\):
\[
(-29)^0 = 1
\]

4. Substitute this value back into the expression:
\[
\frac{1}{2}(9 - 1) + 1
\]

5. Simplify \(9 - 1\):
\[
9 - 1 = 8
\]

6. Now the expression becomes:
\[
\frac{1}{2}(8) + 1
\]

7. Calculate \(\frac{1}{2}(8)\):
\[
\frac{1}{2}(8) = 4
\]

8. Now combine this with the 1:
\[
4 + 1 = 5
\]

So, the value of the left side is \(5\).

**Now, evaluating the right side:**

\[
\frac{1}{2}(9 - 1) - 1
\]

1. Simplify \(9 - 1\):
\[
9 - 1 = 8
\]

2. Now the expression becomes:
\[
\frac{1}{2}(8) - 1
\]

3. Calculate \(\frac{1}{2}(8)\):
\[
\frac{1}{2}(8) = 4
\]

4. Now subtract 1 from this:
\[
4 - 1 = 3
\]

So, the value of the right side is \(3\).

**Final Comparison:**

The left side evaluates to \(5\) and the right side evaluates to \(3\). Thus, the two expressions are not equivalent:

\[
\frac{1}{2}(9 - 7^0) + (-29)^0 \neq \frac{1}{2}(9 - 1) - 1
\]

In conclusion, the statement is false; the two sides are not equivalent.