MAT101 MOD3 Case - 5000= 2500e^0.03t 2 = e^0.03t ln2 =...

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RODNEY BUCKMIRE MAT101 – Module 3 – CASE Dr. Schek 1. F= A(1=r)^n F= 1000(1=0.09)^8 F=1000(1.09)^8 F= $1992.56 2. F(2)=3^2 f(2)=9 3. 1296^x = 6 log1296^x=log 6 x= log6/log1296 x- 0.25 pr ¼ 4. F= Ae^rt
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Unformatted text preview: 5000= 2500e^0.03t 2 = e^0.03t ln2 = 0.03tlne ln2== 0.03t t= ln2/0.03 t = 23.1 years 5. 3 log (6x-4 3log(6x-4)/log3 3log[2?(3x-2)]/log3 3log2/log3 + 3log(3x-2)/log3...
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This note was uploaded on 10/26/2011 for the course MAT 101 taught by Professor Choi during the Fall '11 term at Trident Technical College.

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