Question

A company deposits $25,000 in a bank account that earns interest. The value of the account can be represented as 25,000e^0.0198t, where t represents time, in years.

The time, t, in years that it will take for the account to have a value of $30,000 can be modeled as t=ln(k)/0.0198. What is the value of k?

Answer 1

k = 1.2

Explanation:

Solving for t when the account value is 30,000, we have ...

30,000 = 25,000e^(0.0198t)

30,000/25,000 = 1.2 = e^(0.0198t)

ln(1.2) = 0.0198t

t = ln(1.2)/0.0198

Matching this with the form ...

t = ln(k)/0.0198

we see that k = 1.2.

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k is the multiplier of account value in time period t.