Which table represents the graph of a logarithmic function in the form [tex]$y=\log _b x$[/tex] when [tex]$b \ \textgreater \ 1$[/tex]?

\begin{tabular}{|c|c|}
\hline [tex]$x$[/tex] & [tex]$y$[/tex] \\
\hline[tex]$\frac{1}{8}$[/tex] & -3 \\
\hline[tex]$\frac{1}{4}$[/tex] & -2 \\
\hline[tex]$\frac{1}{2}$[/tex] & -1 \\
\hline 1 & 0 \\
\hline 2 & 1 \\
\hline
\end{tabular}

\begin{tabular}{|c|c|}
\hline [tex]$x$[/tex] & [tex]$y$[/tex] \\
\hline-1.9 & -2.096 \\
\hline 1.75 & -1.262 \\
\hline
\end{tabular}



Answer :

To determine the base [tex]\( b \)[/tex] of the logarithmic function [tex]\( y = \log_b x \)[/tex] from the given tables, follow these steps:

1. Understanding the Function:
- The logarithmic function is given as [tex]\( y = \log_b x \)[/tex]. This means that [tex]\( b^y = x \)[/tex]. Our goal is to find the base [tex]\( b \)[/tex].

2. Using a Known Point:
- Take the point [tex]\((x, y) = (2, 1)\)[/tex] from the table.
- According to the equation [tex]\( y = \log_b x \)[/tex], substituting [tex]\( x = 2 \)[/tex] and [tex]\( y = 1 \)[/tex], we get:
[tex]\[ \log_b 2 = 1 \][/tex]
- This implies that [tex]\( b^1 = 2 \)[/tex], or simply [tex]\( b = 2 \)[/tex].

3. Verification with Another Point:
- Let's verify this base [tex]\( b \)[/tex] using another point. Consider [tex]\((x, y) = \left(\frac{1}{2}, -1\right)\)[/tex].
- Substituting these values into the equation [tex]\( y = \log_b x \)[/tex] gives:
[tex]\[ \log_b \left(\frac{1}{2}\right) = -1 \][/tex]
- This implies that [tex]\( b^{-1} = \frac{1}{2} \)[/tex], or equivalently [tex]\( \frac{1}{b} = \frac{1}{2} \)[/tex].
- Solving this, we get [tex]\( b = 2 \)[/tex].

4. Final Verification and Decision:
- To double-check the correctness, consider that all pairs [tex]\((x, y)\)[/tex] must satisfy [tex]\( y = \log_b x \)[/tex] using [tex]\( b = 2 \)[/tex]. Verifying for a few more points from the table will confirm our result:
- For [tex]\((x, y) = \left(\frac{1}{4}, -2\right)\)[/tex]:
[tex]\[ \log_2 \left(\frac{1}{4}\right) = -2 \implies 2^{-2} = \frac{1}{4} \][/tex]
- For [tex]\((x, y) = (1, 0)\)[/tex]:
[tex]\[ \log_2 (1) = 0 \implies 2^0 = 1 \][/tex]

Since these points verify correctly using [tex]\( b = 2 \)[/tex], the base [tex]\( b \)[/tex] of the logarithmic function [tex]\( y = \log_b x \)[/tex] is indeed 2.

Thus, the derived base [tex]\( b \)[/tex] for the function [tex]\( y = \log_b x \)[/tex] that fits the given values is:
[tex]\[ \boxed{2} \][/tex]