Answer :
To approximate [tex]$\log_5 11$[/tex], we can use the change of base formula, which states that for any positive numbers [tex]$a$[/tex], [tex]$b$[/tex] (with [tex]$b \neq 1$[/tex]), and any base [tex]$c > 0$[/tex] (with [tex]$c \neq 1$[/tex]),
[tex]$$
\log_b a = \frac{\ln a}{\ln b}.
$$[/tex]
Here, we want to find [tex]$\log_5 11$[/tex]. So, we write
[tex]$$
\log_5 11 = \frac{\ln 11}{\ln 5}.
$$[/tex]
The natural logarithm (base [tex]$e$[/tex]) of [tex]$11$[/tex] is approximately [tex]$2.3979$[/tex], and the natural logarithm of [tex]$5$[/tex] is approximately [tex]$1.6094$[/tex]. Dividing these gives
[tex]$$
\log_5 11 \approx \frac{2.3979}{1.6094} \approx 1.4899.
$$[/tex]
Rounding [tex]$1.4899$[/tex] to two decimal places yields
[tex]$$
\log_5 11 \approx 1.49.
$$[/tex]
Thus, the approximate value of [tex]$\log_5 11$[/tex] is [tex]$\boxed{1.49}$[/tex].
[tex]$$
\log_b a = \frac{\ln a}{\ln b}.
$$[/tex]
Here, we want to find [tex]$\log_5 11$[/tex]. So, we write
[tex]$$
\log_5 11 = \frac{\ln 11}{\ln 5}.
$$[/tex]
The natural logarithm (base [tex]$e$[/tex]) of [tex]$11$[/tex] is approximately [tex]$2.3979$[/tex], and the natural logarithm of [tex]$5$[/tex] is approximately [tex]$1.6094$[/tex]. Dividing these gives
[tex]$$
\log_5 11 \approx \frac{2.3979}{1.6094} \approx 1.4899.
$$[/tex]
Rounding [tex]$1.4899$[/tex] to two decimal places yields
[tex]$$
\log_5 11 \approx 1.49.
$$[/tex]
Thus, the approximate value of [tex]$\log_5 11$[/tex] is [tex]$\boxed{1.49}$[/tex].
