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If $3000 is invested in an account earning 11% annual interest compounded continuously, then the number of years that In 2 0.11 Find the number of years. it takes for the amount to grow to $6000 is n = The approximate number of years for the amount to grow to $6000 is years. (Do not round until the final answer. Then round to the nearest tenth as needed.)

If $3000 is invested in an account earning 11% annual interest compounded continuously, then the number of years that In 2 0.11 Find the number of years. it takes for the amount to grow to $6000 is n = The approximate number of years for the amount to grow to $6000 is years. (Do not round until the final answer. Then round to the nearest tenth as needed.)

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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If $3000 is invested in an account earning 11% annual interest compounded continuously, then the number of years that In 2 0.11 Find the number of years. it takes for the amount to grow to $6000 is n = The approximate number of years for the amount to grow to $6000 is years. (Do not round until the final answer. Then round to the nearest tenth as needed.)
Transcribed Image Text:If $3000 is invested in an account earning 11% annual interest compounded continuously, then the number of years that In 2 0.11 Find the number of years. it takes for the amount to grow to $6000 is n = The approximate number of years for the amount to grow to $6000 is years. (Do not round until the final answer. Then round to the nearest tenth as needed.)
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