We have the equation:
[tex]y=\log_3x.[/tex]a) To express x in terms of y, we take into account the following property of logarithms:
[tex]y=\log_ax\Rightarrow x=a^y.[/tex]Choosing a = 3, we have:
[tex]x=3^y.[/tex]b) Now, we evaluate this function for the values y = -2, -1, 0, 1 and 2. To do that, we simply replace the value of y in the last equation, then we simplify the result:
1) y = -2
Replacing y = -2 in the equation of x, we have:
[tex]x=3^{-2}.[/tex]Now, the negative power of a number can be written as the positive power but in the denominator:
[tex]x=\frac{1}{3^2}.[/tex]The 3 square is equal to 3*3 = 9, so we have:
[tex]x=\frac{1}{9}.[/tex]So the point for y = -2 is:
[tex](x,y)=(\frac{1}{9},-2).[/tex]2) y = -1
[tex]y=-1\operatorname{\Rightarrow}x=3^{-1}=\frac{1}{3^1}=\frac{1}{3}\Rightarrow(x,y)=(\frac{1}{3},-1).[/tex]3) y = 0
[tex]y=0\operatorname{\Rightarrow}x=3^0=1\Rightarrow(x,y)=(1,0).[/tex]4) y = 1
[tex]y=1\operatorname{\Rightarrow}x=3^1=3\Rightarrow(x,y)=(3,1).[/tex]5) y = 2
[tex]y=2\Rightarrow x=3^2=9\Rightarrow(x,y)=(9,2).[/tex]Answera) The exponential form is: x = 3^y
b) The values of x for the table are:
• y = -2 ⇒ (,1/9,, -2)
,• y = -1 ⇒ (,1/3,, -1)
,• y = 0 ⇒ (,1,, 0)
,• y = 1 ⇒ (,3,, 1)
• y = 2 ⇒ (,9,, 2)