Since \( e^{1} \approx 2.718 \) and \( e^{2} \approx 7.389 \), betweet what two consecutive imerets is \( \ln 5.6 \) ? A. 5 and 6 B. 2 and 3 C. 1 and 2 D. 0 and 1 Concepr Check Without usinig a calculator, afve the value of long 10 rot Concept Check Without using a calculator, give the value of \( \ln e^{v 1} \)

To determine between which two consecutive integers \( \ln 5.6 \) falls, we can use the approximations of \( e^1 \approx 2.718 \) and \( e^2 \approx 7.389 \). Since \( 5.6 \) is greater than \( 2.718 \) and less than \( 7.389 \), it follows that \( \ln 5.6 \) is between \( 1 \) and \( 2 \). Therefore, the answer is **C. 1 and 2**. For the concept check regarding \( \ln 10 \), we know that \( e^2 \approx 7.389 \) and \( e^3 \approx 20.085 \). Since \( 10 \) is between these two values, we expect \( \ln 10 \) to fall between \( 2 \) and \( 3 \). As for \( \ln e^{v 1} \), using the properties of logarithms, we can simplify it to \( v 1 \), since \( \ln e^x = x \). So, \( \ln e^{v 1} = v 1 \).