The amount of money invested in a certain account increases according to the following function, where $y_{0}$ is the initial amount of the investment, and $y$ is the amount present at time $t$ (in years). $\newline$ $y=y_{0} e^{0.015 t}$ $\newline$ After how many years will the initial investment be doubled? Do not round any intermediate computations, and round your answer to the nearest tenth.

Full solution

Q. The amount of money invested in a certain account increases according to the following function, where

y_{0}

is the initial amount of the investment, and

y

is the amount present at time

t

(in years).

\newline

y=y_{0} e^{0.015 t}

\newline

After how many years will the initial investment be doubled? Do not round any intermediate computations, and round your answer to the nearest tenth.

Set Equation for Doubling: To find when the investment doubles, set $y$ to $2y_{0}$ because that's double the initial amount. $\newline$ So, $2y_{0} = y_{0}e^{0.015 t}$ .
Divide and Simplify: Divide both sides by $y_{0}$ to get $2 = e^{0.015 t}$ .
Take Natural Logarithm: Take the natural logarithm ( $\ln$ ) of both sides to solve for $t$ . $\ln(2) = \ln(e^{0.015 t})$ .
Apply Logarithm Property: Using the property of logarithms, bring down the exponent: $\ln(2) = 0.015 t \times \ln(e)$ .
Solve for $t$ : Since $\ln(e) = 1$ , this simplifies to $\ln(2) = 0.015 t$ .
Calculate $t$ : Divide both sides by $0.015$ to isolate $t$ . $\newline$ $t = \frac{\ln(2)}{0.015}$ .
Round to Nearest Tenth: Calculate $t$ using a calculator. $\newline$ $t \approx \frac{\ln(2)}{0.015} \approx 46.2057$ .
Round to Nearest Tenth: Calculate $t$ using a calculator. $t \approx \ln(2) / 0.015 \approx 46.2057$ .Round the answer to the nearest tenth. $t \approx 46.2$ years.