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If 8000 dollars is invested in a bank account at an interest rate of 5% per year, compounded continuously? How many years will it take for your balance to reach 20000 dollars?
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y=Ae^(kt)
k=0.05 , the "growth constant" , which is 5%
The initial value is 8000 , so A=8000 \
y=8000e^(0.05t)
20 000 = 8000e^(0.05t) Solve for t.
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e^(0.05t) = (20/8) 0.05t = ln(20/8) t= (1/0.05) ln(20/8)
\[t= \frac{1}{0.05} \ln(\frac{20}{8})\] and that is in years. I don't have a calculator but you can sub into a calculator to get the actual value.
" compounded continuously" means it is an exponential growth,
thank you very much
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