Lab 12 Solutions

Fall 2023 MATH 1018: Pre-Calculus in Practice Lab 12 More Exponential Functions and Logarithmic Functions Name & Student ID # : Name & Student ID # : Name & Student ID # : Name & Student ID # : Instructions • Form groups of 1–4 and write the group-members names in the space above. Only one submission per group is allowed. All group members should agree on an answer before it is written down. Groups can change from week to week. All group members will receive the same mark. • Make sure to write down the steps of your work unless otherwise instructed. Final answers without the steps shown are not sufficient to earn full completion credit. • You may choose to use a calculator for any lab problems. However you are encouraged not to use a calculator for most lab problems since you will not be allowed to use a calculator during quizzes or the final exam. Intended Learning Outcomes In this lab, you will be practicing the following skills in a modeling setting. • Solve simple equations involving exponential and logarithmic expressions • Graph logarithmic functions • Work with common models build on exponential functions MATH 1018: Pre-Calculus in Practice Page 1

Examples 1. Tony invested $40000 at the beginning of 2010. Unfortunately his investment has been losing value at the rate of 2.5% per year. (a) Find a formula for the value of his investment as a function of time. Solution: Let V ( t ) be the value of Tony’s investment t years after the beginning of 2010. Then V ( t ) = ab t for some a ̸ = 0 , b > 0 , b ̸ = 1 . The initial investment was $40000, so a = 40000 . Since the investment loses 2.5% of its value each year, its value would get multiplied by a factor of 0.975 each year, so b = 0 . 975 . Therefore V ( t ) = 40000(0 . 975) t . (b) Change the base of the exponential function found in part (a) to e . Base on the result, if we consider the decrease of value as a continuous process, what would the continuous decay rate be? Solution: We are trying change the base of the exponential function V ( t ) , so we want to rewrite it as V ( t ) = 40000 e rt for some value r instead. Then we need to convert 0 . 975 to a base- e number first. Note that 0 . 975 = e ln(0 . 975) . Therefore, V ( t ) = 40000 e ln(0 . 975) t . The continuous decay rate is − ln(0 . 975) , or approximately 2 . 53% . (c) Now suppose Steve invested $40000 at the beginning of 2015. This investment has also been decreasing in value at the rate of 3.1% per year. When will the value Tony’s invest- ment and Steve’s investment equal in value? Solution: Recall that back in part (a), t was defined as the number of years after 2010. Here we use the same definition of t , then by the same logic, the value of Steve’s investment, S ( t ) , can be modelled as S ( t ) = 40000(0 . 969) t − 5 , for t ≥ 5 . Then we are trying to solve for t when 40000(0 . 975) t = 40000(0 . 969) t − 5 . MATH 1018: Pre-Calculus in Practice Page 2