Question

Use the equation and type the ordered-pairs. y = log 3 x {(1/3, a0), (1, a1), (3, a2), (9, a3), (27, a4), (81, a5)

Answer 1

Answer:

Considering the given equation $y = log_{3}x$

And the ordered pairs in the format $(x, y)$

I don't know if it is log of base 3 or 10, but I will assume it is 3.

For $(\frac{1}{3}, a_{0} )$

$x=\frac{1}{3}$

$y=a_{0}$

$y = log_{3}x\\y = log_{3}(\frac{1}{3} )\\y=-\log _3\left(3\right)\\y=-1$

So the ordered pair will be $(\frac{1}{3}, -1 )$

For $(1, a_{1} )$

$x=1$

$y=a_{1}$

$y = log_{3}x\\y = log_{3}1\\y = log_{3}(1)\\Note: \log _a(1)=0\\y = 0$

So the ordered pair will be $(1, 0 )$

For $(3, a_{2} )$

$x=3$

$y=a_{2}$

$y = log_{3}x\\y = log_{3}3\\y = 1$

So the ordered pair will be $(3, 1 )$

For $(9, a_{3} )$

$x=9$

$y=a_{3}$

$y = log_{3}x\\y = log_{3}9\\y=2\log _3\left(3\right)\\y=2$

So the ordered pair will be $(9, 2 )$

For $(27, a_{4} )$

$x=27$

$y=a_{4}$

$y = log_{3}x\\y = log_{3}27\\y=3\log _3\left(3\right)\\y=3$

So the ordered pair will be $(27, 3 )$

For $(81, a_{5} )$

$x=81$

$y=a_{5}$

$y = log_{3}x\\y = log_{3}81\\y=4\log _3\left(3\right)\\y=4$

So the ordered pair will be $(81, 4 )$