which student has correctly rewritten \\(\\log_{6} 1.1\\) using the change of base formula?\\nstudent #1…

which student has correctly rewritten \\(\\log_{6} 1.1\\) using the change of base formula?\\nstudent #1: \\(\\frac{\\ln 1.1}{\\ln 6}\\)\\nstudent #2: \\(\\log \\frac{1.1}{6}\\)\\nstudent #3: \\(\\frac{\\log 6}{\\log 1.1}\\)\\n(1 point)\\nstudent #\\(\\square\\)

which student has correctly rewritten \\(\\log_{6} 1.1\\) using the change of base formula?\\nstudent #1: \\(\\frac{\\ln 1.1}{\\ln 6}\\)\\nstudent #2: \\(\\log \\frac{1.1}{6}\\)\\nstudent #3: \\(\\frac{\\log 6}{\\log 1.1}\\)\\n(1 point)\\nstudent #\\(\\square\\)

Answer

Explanation:

Step1: Recall change of base formula

The change of base formula for logarithms is $\log_b a = \frac{\log_k a}{\log_k b}$ (where $k>0, k\neq1$; this includes natural logs $\ln$, where $k=e$).

Step2: Apply formula to $\log_6 1.1$

Substitute $b=6$, $a=1.1$. Using natural log ($k=e$), we get $\log_6 1.1 = \frac{\ln 1.1}{\ln 6}$.

Step3: Match to student answers

Compare with each student's work: Student #1's expression matches the correct application, Student #2 incorrectly applies a quotient rule instead of change of base, Student #3 reverses the numerator and denominator.

Answer:

Student #1: $\frac{\ln 1.1}{\ln 6}$