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Example 9 In a bank, principal increases continuously at the rate of 5% per year. In how many years Rs 1000 double itself ?Rate of change of principal = 5% of Principal ๐’…๐‘ท/๐’…๐’• = 5% ร— P ๐‘‘๐‘ƒ/๐‘‘๐‘ก = 5/100 ร— P ๐‘‘๐‘ƒ/๐‘‘๐‘ก = 1/20 ร— P ๐’…๐’‘/๐‘ท = ๐’…๐’•/๐Ÿ๐ŸŽ Integrating both sides โˆซ1โ–’๐‘‘๐‘/๐‘ƒ = โˆซ1โ–’๐‘‘๐‘ก/20 log |๐‘ท| = ๐’•/๐Ÿ๐ŸŽ + C Removing log P = e^(๐‘ก/20 + ๐ถ) ร— ec P = e^(๐‘ก/20 + ๐ถ) ร— ec P = k๐’†^(๐’•/๐Ÿ๐ŸŽ) where k = ec Now, we have to find in how many years Rs 1000 double it self Thus, we need to find time T when Principal is Rs 2000 First let us find k At t = 0, P = 1000 Putting in (1) P = ke^(๐‘ก/20) 1000 = k๐’†^(๐ŸŽ/๐Ÿ๐ŸŽ) 1000 = k e0 1000 = k ร— 1 1000 = k So, k = 1000 Put k = 1000 in (1) P = ke^(๐‘ก/20) P = 1000 ๐’†^(๐’•/๐Ÿ๐ŸŽ) Now, we need to find time t when Principal is Rs 2000 Putting P = 2000, t = t 2000 = 1000 e^(๐‘ก/20) 2000/1000 = e^(๐‘ก/20) 2 = e^(๐‘ก/20) ๐’†^(๐’•/๐Ÿ๐ŸŽ) = 2 Taking log both sides ใ€–๐ฅ๐จ๐ ใ€—_๐’†โกใ€–๐’†^(๐’•/๐Ÿ๐ŸŽ) ใ€—=ใ€–๐ฅ๐จ๐ ใ€—_๐’†โก๐Ÿ t/20 log_๐‘’โก๐‘’=log_๐‘’โก2 " " t/20ร— 1=log_๐‘’โก2 ๐ญ=๐Ÿ๐ŸŽ ใ€–๐’๐’๐’ˆใ€—_๐’†โก๐Ÿ

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo