Answer:
The satisfied table of the given function[tex]y = log_{b} (x)[/tex]
x 1/8 1/4 1/2 1 2
y -3 -2 -1 0 1
Step-by-step explanation:
Explanation :-
Given logarithmic function [tex]y = log_{b} (x)[/tex] if b >1
Given first table
i)
put x = [tex]\frac{1}{8}[/tex] given b > 1 so we can choose b = 2
[tex]y = log_{2} (\frac{1}{8} )[/tex]
[tex]y = log_{2} (2^{-3} )[/tex]
we will apply logarithmic formula
log x ⁿ = n log (x)
[tex]y = log_{2} (2^{-3} ) = -3 log_{2} (2) = -3 (1) = -3[/tex]
y = -3
ii)
put x = [tex]\frac{1}{4}[/tex] given b > 1 so we can choose b = 2
[tex]y = log_{2} (\frac{1}{4} )[/tex]
[tex]y = log_{2} (2^{-2} )[/tex]
we will apply logarithmic formula
log x ⁿ = n log (x)
[tex]y = log_{2} (2^{-2} ) = -2 log_{2} (2) = -2 (1) = -2[/tex]
y = -2
iii)
put x = [tex]\frac{1}{2}[/tex] given b > 1 so we can choose b = 2
[tex]y = log_{2} (\frac{1}{2} )[/tex]
[tex]y = log_{2} (2^{-1} )[/tex]
we will apply logarithmic formula
log x ⁿ = n log (x)
[tex]y = log_{2} (2^{-1} ) = -1 log_{2} (2) = - (1) = -1[/tex]
y = -1
iv)
put x = 1 given b > 1 so we can choose b = 2
[tex]y = log_{2} (1 )[/tex] = 0
y = 0
v)
put x = [tex]2[/tex] given b > 1 so we can choose b = 2
[tex]y = log_{2} (2 )[/tex]
y = 1
Final answer:-
The satisfied table of the given function
x 1/8 1/4 1/2 1 2
y -3 -2 -1 0 1
The first option is correct because that table of values represents the graph of a logarithmic function in the form [tex]y=\log_b x,b>1[/tex].
Given:
The lograrithmic function form is [tex]y=\log_b x,b>1[/tex].
To find:
The table of values that represents the graph of a logarithmic function in the form [tex]y=\log_b x,b>1[/tex].
Explanation:
It is given that [tex]b>1[/tex]. Let [tex]b=2[/tex], then
[tex]y=\log_2 x[/tex] ...(i)
Substitute [tex]x=\dfrac{1}{8}[/tex] in (i), we get
[tex]y=\log_2 \dfrac{1}{8}[/tex]
[tex]y=\log_2 \dfrac{1}{2^3}[/tex]
[tex]y=\log_2 2^{-3}[/tex]
[tex]y=-3[/tex]
Substitute [tex]x=\dfrac{1}{4}[/tex] in (i), we get
[tex]y=\log_2 \dfrac{1}{4}[/tex]
[tex]y=\log_2 \dfrac{1}{2^2}[/tex]
[tex]y=\log_2 2^{-2}[/tex]
[tex]y=-2[/tex]
Substitute [tex]x=\dfrac{1}{2}[/tex] in (i), we get
[tex]y=\log_2 \dfrac{1}{2}[/tex]
[tex]y=\log_2 2^{-1}[/tex]
[tex]y=-1[/tex]
Substitute [tex]x=1[/tex] in (i), we get
[tex]y=\log_2 1[/tex]
[tex]y=0[/tex]
Substitute [tex]x=2[/tex] in (i), we get
[tex]y=\log_2 2[/tex]
[tex]y=1[/tex]
The table of values is:
[tex]x[/tex] [tex]y[/tex]
[tex]\dfrac{1}{8}[/tex] [tex]-3[/tex]
[tex]\dfrac{1}{4}[/tex] [tex]-2[/tex]
[tex]\dfrac{1}{2}[/tex] [tex]-1[/tex]
[tex]1[/tex] [tex]0[/tex]
[tex]2[/tex] [tex]1[/tex]
Therefore, the first option is correct.
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