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Al asignar valores a la función [tex]$f(x)=\log_3(2x)$[/tex], ¿a qué tabla corresponde?

Seleccione una:

a.
\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$F(x)$[/tex] \\
\hline
2 & 2.1 \\
\hline
3 & 4.6 \\
\hline
6 & 5.8 \\
\hline
\end{tabular}

b.
\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$F(x)$[/tex] \\
\hline
2 & 1.26 \\
\hline
3 & 1.63 \\
\hline
6 & 2.26 \\
\hline
\end{tabular}

c.
\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$F(x)$[/tex] \\
\hline
2 & 1.6 \\
\hline
3 & 2.5 \\
\hline
6 & 3.7 \\
\hline
\end{tabular}

d.
\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$F(x)$[/tex] \\
\hline
2 & 1.001 \\
\hline
3 & 2.2 \\
\hline
6 & 3.4 \\
\hline
\end{tabular}


Sagot :

Para responder a esta pregunta, debemos evaluar la función [tex]\( f(x) = \log_3(2x) \)[/tex] en los valores [tex]\( x \)[/tex] proporcionados (2, 3 y 6) y verificar a cuál de las tablas dadas corresponden los resultados obtenidos.

Paso 1: Evaluación de la función [tex]\( f(x) \)[/tex]

La función [tex]\( f(x) \)[/tex] se puede reescribir utilizando el cambio de base:
[tex]\[ f(x) = \log_3(2x) = \frac{\log_2(2x)}{\log_2(3)} \][/tex]

Paso 2: Cálculo de [tex]\( f(2) \)[/tex]

[tex]\[ f(2) = \frac{\log_2(2 \times 2)}{\log_2(3)} = \frac{\log_2(4)}{\log_2(3)} \][/tex]
Sabemos que:
[tex]\[ \log_2(4) = 2 \][/tex]
Entonces:
[tex]\[ f(2) = \frac{2}{\log_2(3)} \approx 1.26 \][/tex]

Paso 3: Cálculo de [tex]\( f(3) \)[/tex]

[tex]\[ f(3) = \frac{\log_2(2 \times 3)}{\log_2(3)} = \frac{\log_2(6)}{\log_2(3)} \][/tex]
Sabemos que:
[tex]\[ \log_2(6) \approx 2.58496 \][/tex]
Entonces:
[tex]\[ f(3) = \frac{2.58496}{\log_2(3)} \approx 1.63 \][/tex]

Paso 4: Cálculo de [tex]\( f(6) \)[/tex]

[tex]\[ f(6) = \frac{\log_2(2 \times 6)}{\log_2(3)} = \frac{\log_2(12)}{\log_2(3)} \][/tex]
Sabemos que:
[tex]\[ \log_2(12) \approx 3.58496 \][/tex]
Entonces:
[tex]\[ f(6) = \frac{3.58496}{\log_2(3)} \approx 2.26 \][/tex]

Paso 5: Comparación con las tablas proporcionadas

[tex]\[ \begin{aligned} &\text{Tabla a:} \ \begin{tabular}{|c|c|} \hline$x$ & $F(x)$ \\ \hline 2 & 2.1 \\ \hline 3 & 4.6 \\ \hline 6 & 5.8 \\ \hline \end{tabular} \\ &\text{Tabla b:} \ \begin{tabular}{|c|c|} \hline$x$ & $F(x)$ \\ \hline 2 & 1.26 \\ \hline 3 & 1.63 \\ \hline 6 & 2.26 \\ \hline \end{tabular} \\ &\text{Tabla c:} \ \begin{tabular}{|c|c|} \hline$x$ & $F(x)$ \\ \hline 2 & 1.6 \\ \hline 3 & 2.5 \\ \hline 6 & 3.7 \\ \hline \end{tabular} \\ &\text{Tabla d:} \ \begin{tabular}{|c|c|} \hline$x$ & $F(x)$ \\ \hline 2 & 1.001 \\ \hline 3 & 2.2 \\ \hline 6 & 3.4 \\ \hline \end{tabular} \\ \end{aligned} \][/tex]

Comparando los valores calculados [tex]\( f(2) \approx 1.26 \)[/tex], [tex]\( f(3) \approx 1.63 \)[/tex] y [tex]\( f(6) \approx 2.26 \)[/tex] con las tablas proporcionadas, vemos que coinciden con la Tabla b.

Por lo tanto, la función [tex]\( f(x) = \log_3(2x) \)[/tex] corresponde a la Tabla b.
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