Answer:
As a logarithm, the money had been in the account for [tex]t = \frac{\ln{2}}{0.08}[/tex] years.
Step-by-step explanation:
The amount of money in the account after t years is given by:
[tex]A(t) = Pe^{rt}[/tex]
In which P is the initial amount deposited and r is the interest rate, as a decimal.
Michelle deposited $1,600 into an account that is compounded continuously with an annual interest rate of 8%.
This means that [tex]P = 1600, r = 0.08[/tex]
After t years, there was $3,200 in the account. How many years had the money been in the account?
This is t for which [tex]P(t) = 3200[/tex]. So
[tex]A(t) = Pe^{rt}[/tex]
[tex]3200 = 1600e^{0.08t}[/tex]
[tex]e^{0.08t} = \frac{3200}{1600}[/tex]
[tex]e^{0.08t} = 2[/tex]
[tex]\ln{e^{0.08t}} = \ln{2}[/tex]
[tex]0.08t = \ln{2}[/tex]
[tex]t = \frac{\ln{2}}{0.08}[/tex]
[tex]t = 8.67[/tex]
As a logarithm, the money had been in the account for [tex]t = \frac{\ln{2}}{0.08}[/tex] years.
The value of time t is [tex]\frac{ln(2)}{0.08}[/tex].
Given: P=1600
r=8%=0.08
t = ?
A=3200
Using the given compound interest formula we get:
[tex]A=Pe^{rt}\\3200=1600e^{0.08t}\\2=e^{0.08t}\\[/tex]
Take ln on both sides,
[tex]ln(2)=ln(e^{0.08t})\\ln(2)=0.08t\\t=\frac{ln(2)}{0.08}[/tex]
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