Answer :
To ascertain which table correctly represents the graph of a logarithmic function in the form [tex]\( y = \log_8 x \)[/tex] with [tex]\( b > 1 \)[/tex], let’s analyze the values in the tables.
Table 1:
[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $y$ \\ \hline $\frac{1}{8}$ & -3 \\ \hline $\frac{1}{4}$ & -2 \\ \hline $\frac{1}{2}$ & -1 \\ \hline 1 & 0 \\ \hline 2 & 1 \\ \hline \end{tabular} \][/tex]
Table 2:
[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $y$ \\ \hline -1.9 & -2.096 \\ \hline -1.75 & -1.262 \\ \hline \end{tabular} \][/tex]
Breakdown of Table 1:
For [tex]\( y = \log_8 x \)[/tex]:
- When [tex]\( x = 1 \)[/tex]:
[tex]\[ y = \log_8 (1) = 0 \][/tex]
This matches the values [tex]\( (1, 0) \)[/tex] in Table 1.
- When [tex]\( x = \frac{1}{2} \)[/tex]:
- We know [tex]\( 8^{-1} = \frac{1}{8} \)[/tex].
- Therefore: [tex]\( \log_8 \left( \frac{1}{8} \right) = -1 \times \log_8 (8) = -1 \)[/tex].
[tex]\[ y = -1 \][/tex]
Again, this matches the value [tex]\( (\frac{1}{8}, -3) \rightarrow (\frac{1}{2}, -1) \)[/tex].
- When [tex]\( x = 2 \)[/tex]:
- [tex]\( 8^y = 2 \implies y = 1 \)[/tex].
[tex]\[ y = 1 \][/tex]
This matches the value [tex]\( (2, 1) \)[/tex].
By analyzing the values, the function correctly represents the form [tex]\( y = \log_8 x \)[/tex] with [tex]\( b > 1 \)[/tex].
Table 2 Breakdown:
Values provided in Table 2 include negative [tex]\( x \)[/tex] values such as -1.9 and -1.75, but the logarithmic function [tex]\( y = \log_8 x \)[/tex] with base [tex]\( b > 1 \)[/tex] must take a positive [tex]\( x \)[/tex] value.
So, Table 1 correctly represents the logarithmic function [tex]\( y = \log_8 x \)[/tex].
Table 1:
[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $y$ \\ \hline $\frac{1}{8}$ & -3 \\ \hline $\frac{1}{4}$ & -2 \\ \hline $\frac{1}{2}$ & -1 \\ \hline 1 & 0 \\ \hline 2 & 1 \\ \hline \end{tabular} \][/tex]
Table 2:
[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $y$ \\ \hline -1.9 & -2.096 \\ \hline -1.75 & -1.262 \\ \hline \end{tabular} \][/tex]
Breakdown of Table 1:
For [tex]\( y = \log_8 x \)[/tex]:
- When [tex]\( x = 1 \)[/tex]:
[tex]\[ y = \log_8 (1) = 0 \][/tex]
This matches the values [tex]\( (1, 0) \)[/tex] in Table 1.
- When [tex]\( x = \frac{1}{2} \)[/tex]:
- We know [tex]\( 8^{-1} = \frac{1}{8} \)[/tex].
- Therefore: [tex]\( \log_8 \left( \frac{1}{8} \right) = -1 \times \log_8 (8) = -1 \)[/tex].
[tex]\[ y = -1 \][/tex]
Again, this matches the value [tex]\( (\frac{1}{8}, -3) \rightarrow (\frac{1}{2}, -1) \)[/tex].
- When [tex]\( x = 2 \)[/tex]:
- [tex]\( 8^y = 2 \implies y = 1 \)[/tex].
[tex]\[ y = 1 \][/tex]
This matches the value [tex]\( (2, 1) \)[/tex].
By analyzing the values, the function correctly represents the form [tex]\( y = \log_8 x \)[/tex] with [tex]\( b > 1 \)[/tex].
Table 2 Breakdown:
Values provided in Table 2 include negative [tex]\( x \)[/tex] values such as -1.9 and -1.75, but the logarithmic function [tex]\( y = \log_8 x \)[/tex] with base [tex]\( b > 1 \)[/tex] must take a positive [tex]\( x \)[/tex] value.
So, Table 1 correctly represents the logarithmic function [tex]\( y = \log_8 x \)[/tex].
