Certainly! Let's work through each part of the question, where we will change the position of the numerator and denominator.
### Part (a)
Given expression: [tex]\(\frac{3^{-9}}{1}\)[/tex]
Due to the properties of exponents, when we move a term with a negative exponent from the numerator to the denominator (or vice versa), the sign of the exponent changes:
[tex]\[
\frac{3^{-9}}{1} = 3^{-9} = \frac{1}{3^9}
\][/tex]
Calculating [tex]\(3^9\)[/tex], we get:
[tex]\[
3^9 = 19683
\][/tex]
Thus:
[tex]\[
\frac{1}{3^9} = \frac{1}{19683}
\][/tex]
The numerical result is:
[tex]\[
\frac{1}{19683} \approx 5.080526342529086 \times 10^{-5}
\][/tex]
### Part (b)
Given expression: [tex]\(\frac{a^m}{b^m}\)[/tex]
Using the property of exponents [tex]\(\frac{a^m}{b^m} = (\frac{a}{b})^m\)[/tex]:
[tex]\[
\frac{a^m}{b^m} = \left(\frac{a}{b}\right)^m
\][/tex]
For example, if [tex]\(a = 2\)[/tex], [tex]\(b = 3\)[/tex], and [tex]\(m = 4\)[/tex]:
[tex]\[
\frac{2^4}{3^4} = \left(\frac{2}{3}\right)^4
\][/tex]
Calculating [tex]\((\frac{2}{3})^4\)[/tex]:
[tex]\[
\left(\frac{2}{3}\right)^4 = \frac{2^4}{3^4} = \frac{16}{81}
\][/tex]
The numerical result is:
[tex]\[
\frac{16}{81} \approx 0.19753086419753085
\][/tex]
### Part (c)
Given expression: [tex]\(\frac{p^{-8}}{p}\)[/tex]
Using the property of exponents, and the rule [tex]\(\frac{p^{-8}}{p} = p^{-8 - 1} = p^{-9}\)[/tex]:
[tex]\[
\frac{p^{-8}}{p} = p^{-9} = \frac{1}{p^9}
\][/tex]
For example, if [tex]\(p = 5\)[/tex]:
[tex]\[
p^{-9} = \frac{1}{5^9}
\][/tex]
Calculating [tex]\(5^9\)[/tex], we get:
[tex]\[
5^9 = 1953125
\][/tex]
Thus:
[tex]\[
\frac{1}{5^9} = \frac{1}{1953125}
\][/tex]
The numerical result is:
[tex]\[
\frac{1}{1953125} \approx 5.12 \times 10^{-7}
\][/tex]
To summarize, we have the solutions to each part when the numerators are changed into denominators:
- (a) [tex]\(\frac{1}{19683} \approx 5.080526342529086 \times 10^{-5}\)[/tex]
- (b) [tex]\(\left(\frac{2}{3}\right)^4 = 0.19753086419753085\)[/tex]
- (c) [tex]\(\frac{1}{1953125} \approx 5.12 \times 10^{-7}\)[/tex]
