rewrite the expression \\(\frac{9^{-8} cdot 3^{0}}{2^{-1}}\\) using only positive exponents. show your work.

Explanation:

Step1: Apply zero exponent rule

Recall that any non - zero number to the power of 0 is 1, so (3^{0}=1). Also, use the negative exponent rule (a^{-n}=\frac{1}{a^{n}}) and (\frac{1}{a^{-n}} = a^{n}). First, simplify the numerator and the denominator separately. The numerator is (9^{-8}\cdot3^{0}), and since (3^{0} = 1), the numerator becomes (9^{-8}\times1=9^{-8}). The denominator is (2^{-1}), and by the negative exponent rule, (2^{-1}=\frac{1}{2^{1}}), so (\frac{1}{2^{-1}}=2^{1}). Now our expression is (\frac{9^{-8}\times1}{2^{-1}}=\frac{9^{-8}}{2^{-1}}). Using the negative exponent rule on (9^{-8}), we know that (9^{-8}=\frac{1}{9^{8}}), and (\frac{1}{2^{-1}} = 2). So (\frac{9^{-8}}{2^{-1}}=\frac{1}{9^{8}}\times2=\frac{2}{9^{8}}) We can also rewrite (9) as (3^{2}), so (9^{8}=(3^{2})^{8}). By the power of a power rule ((a^{m})^{n}=a^{mn}), ((3^{2})^{8}=3^{16}). Then (\frac{2}{9^{8}}=\frac{2}{3^{16}}) (but (\frac{2}{9^{8}}) is also correct with positive exponents) Another way: Start with the original expression (\frac{9^{-8}\cdot3^{0}}{2^{-1}}) We know that (9 = 3^{2}), so (9^{-8}=(3^{2})^{-8}). By the power of a power rule ((a^{m})^{n}=a^{mn}), ((3^{2})^{-8}=3^{-16}) The expression becomes (\frac{(3^{2})^{-8}\cdot3^{0}}{2^{-1}}) Using the power of a power rule: ((3^{2})^{-8}=3^{-16}), and (3^{0} = 1), so the numerator is (3^{-16}\times1 = 3^{-16}) The denominator is (2^{-1}), so the expression is (\frac{3^{-16}}{2^{-1}}) Using the negative exponent rule (\frac{a^{-m}}{b^{-n}}=\frac{b^{n}}{a^{m}}), we get (\frac{2^{1}}{3^{16}}=\frac{2}{3^{16}}) or since (9^{8}=(3^{2})^{8}=3^{16}), we can also write it as (\frac{2}{9^{8}})

Step2: Simplify using exponent rules

• Zero exponent: (a^{0}=1) ((a\neq0)), so (3^{0} = 1)
• Negative exponent: (a^{-n}=\frac{1}{a^{n}}) and (\frac{1}{a^{-n}}=a^{n})
  • For (9^{-8}), (9^{-8}=\frac{1}{9^{8}})
  • For (2^{-1}), (\frac{1}{2^{-1}}=2)

So (\frac{9^{-8}\cdot3^{0}}{2^{-1}}=\frac{9^{-8}\times1}{2^{-1}}=\frac{\frac{1}{9^{8}}}{\frac{1}{2}}=\frac{1}{9^{8}}\times2=\frac{2}{9^{8}}) (or (\frac{2}{3^{16}}) as (9 = 3^{2}), (9^{8}=(3^{2})^{8}=3^{16}))

Answer:

(\frac{2}{9^{8}}) (or (\frac{2}{3^{16}}))