Answer:
Considering the given equation [tex]y = log_{3}x\\[/tex]
And the ordered pairs in the format [tex](x, y)[/tex]
I don't know if it is log of base 3 or 10, but I will assume it is 3.
For [tex](\frac{1}{3}, a_{0} )[/tex]
[tex]x=\frac{1}{3}[/tex]
[tex]y=a_{0}[/tex]
[tex]y = log_{3}x\\y = log_{3}(\frac{1}{3} )\\y=-\log _3\left(3\right)\\y=-1[/tex]
So the ordered pair will be [tex](\frac{1}{3}, -1 )[/tex]
For [tex](1, a_{1} )[/tex]
[tex]x=1[/tex]
[tex]y=a_{1}[/tex]
[tex]y = log_{3}x\\y = log_{3}1\\y = log_{3}(1)\\Note: \log _a(1)=0\\y = 0[/tex]
So the ordered pair will be [tex](1, 0 )[/tex]
For [tex](3, a_{2} )[/tex]
[tex]x=3[/tex]
[tex]y=a_{2}[/tex]
[tex]y = log_{3}x\\y = log_{3}3\\y = 1[/tex]
So the ordered pair will be [tex](3, 1 )[/tex]
For [tex](9, a_{3} )[/tex]
[tex]x=9[/tex]
[tex]y=a_{3}[/tex]
[tex]y = log_{3}x\\y = log_{3}9\\y=2\log _3\left(3\right)\\y=2[/tex]
So the ordered pair will be [tex](9, 2 )[/tex]
For [tex](27, a_{4} )[/tex]
[tex]x=27[/tex]
[tex]y=a_{4}[/tex]
[tex]y = log_{3}x\\y = log_{3}27\\y=3\log _3\left(3\right)\\y=3[/tex]
So the ordered pair will be [tex](27, 3 )[/tex]
For [tex](81, a_{5} )[/tex]
[tex]x=81[/tex]
[tex]y=a_{5}[/tex]
[tex]y = log_{3}x\\y = log_{3}81\\y=4\log _3\left(3\right)\\y=4[/tex]
So the ordered pair will be [tex](81, 4 )[/tex]
