Question

what happens with zero and negative exponents?
a. complete the table to show what it means to have an exponent of zero or a negative exponent.
b. as you move toward the left, each number is being multiplied by 2. what is the multiplier as you move toward the right? hint: what do you do to go from 8 → 4 → 2 → 1?
c. use the patterns you found in the table to write 2^(-6) as a fraction. hint: 2^(-a)=1/2^a
d. write 1/32 as a power of 2 with a single exponent.
e. what is the value of 2^0? hint: any number to the 0 power is...

Explanation:

Step1: Recall exponent rules

For any non - zero number \(a\), \(a^0 = 1\) and \(a^{-n}=\frac{1}{a^{n}}\), where \(n\) is a positive integer.

Step2: Complete part a (assuming a pattern of powers of 2)

If we start with \(2^4 = 16\), then moving left: \(2^3=8\), \(2^2 = 4\), \(2^1=2\), \(2^0 = 1\), \(2^{-1}=\frac{1}{2}\), \(2^{-2}=\frac{1}{4}\), \(2^{-3}=\frac{1}{8}\), \(2^{-4}=\frac{1}{16}\)

Step3: Solve part b

To go from \(8
ightarrow4
ightarrow2
ightarrow1\), we divide by 2 each time. So the multiplier as we move to the right is \(\frac{1}{2}\)

Step4: Solve part c

Using the rule \(a^{-n}=\frac{1}{a^{n}}\), for \(a = 2\) and \(n = 6\), \(2^{-6}=\frac{1}{2^{6}}=\frac{1}{64}\)

Step5: Solve part d

Since \(32=2^5\), then \(\frac{1}{32}=2^{-5}\)

Step6: Solve part e

By the zero - exponent rule, for any non - zero number \(a\), \(a^0 = 1\). So \(2^0=1\)

Answer:

a. (Pattern of powers of 2 as described above)
b. \(\frac{1}{2}\)
c. \(\frac{1}{64}\)
d. \(2^{-5}\)
e. \(1\)