2. The value of a car depreciates exponentially over time. The function v = 26,500(2)-0.18+ can be used to determine v, the value of the car t years after its initial purchase. Which expression represents the number of years that will elapse before the car has a value of $12,000? log (26,500 12,000. 0.18 12,000 log226,500/ 0.18 log 26,500 108₂ 12,000. 0.18 12,000 26,500. 0.18
2. The value of a car depreciates exponentially over time. The function v = 26,500(2)-0.18+ can be used to determine v, the value of the car t years after its initial purchase. Which expression represents the number of years that will elapse before the car has a value of $12,000? log (26,500 12,000. 0.18 12,000 log226,500/ 0.18 log 26,500 108₂ 12,000. 0.18 12,000 26,500. 0.18
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
Related questions
Question
![2. The value of a car depreciates exponentially over time. The function
v = 26,500(2)-0.18t can be used to determine v, the value of the car t years after
its initial purchase. Which expression represents the number of years that will
elapse before the car has a value of $12,000?
log
26,500
12,000.
0.18
12,000
1082 26,500/
0.18
26,500
log2 12,000.
0.18
log
12,000
26,500.
0.18](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9f87d58e-427f-46fd-9431-a8ef93d047c1%2Faa3e58c7-4754-40de-b655-b32213175da2%2Fysipwkd_processed.png&w=3840&q=75)
Transcribed Image Text:2. The value of a car depreciates exponentially over time. The function
v = 26,500(2)-0.18t can be used to determine v, the value of the car t years after
its initial purchase. Which expression represents the number of years that will
elapse before the car has a value of $12,000?
log
26,500
12,000.
0.18
12,000
1082 26,500/
0.18
26,500
log2 12,000.
0.18
log
12,000
26,500.
0.18
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