Full solution

Q. How long will it take

\$ 3,000

to grow to

\$ 10,000

if it is invested at

6 \%

compounded monthly?

\newline

\square

years (Round to the nearest tenth of a year.)

Identify Variables: Identify the variables from the problem. $\newline$ Principal amount $P$ = $\$3,000$ $\newline$ Future value $A$ = $\$10,000$ $\newline$ Annual interest rate $r$ = $6\%$ or $0.06$ (as a decimal) $\newline$ Compounded monthly means the number of compounding periods per year $n$ = $12$ $\newline$ We need to find the time $t$ in years.
Use Formula: Use the compound interest formula to set up the equation. $\newline$ The compound interest formula is $A = P(1 + r/n)^{(nt)}$ . $\newline$ Substitute the given values into the formula to get: $\newline$ $\$$ $10$ , $000$ = $\$$ $3$ , $000$ $(1 + 0.06/12)^{(12t)}$ .
Simplify Equation: Simplify the equation. $\newline$ $\$10,000 = \$3,000(1 + 0.005)^{12t}$ $\newline$ $\$10,000 = \$3,000(1.005)^{12t}$
Isolate Exponential Part: Divide both sides of the equation by $\$3,000$ to isolate the exponential part. $\newline$ $\$10,000 / \$3,000 = (1.005)^{12t}$ $\newline$ $3.3333\ldots = (1.005)^{12t}$
Take Natural Logarithm: Take the natural logarithm ( $\ln$ ) of both sides to solve for the exponent. $\ln(3.3333\ldots) = \ln((1.005)^{12t})$ $\ln(3.3333\ldots) = 12t \cdot \ln(1.005)$
Solve for Exponent: Divide both sides by $12 \times \ln(1.005)$ to solve for $t$ . $\newline$ $t = \frac{\ln(3.3333\ldots)}{12 \times \ln(1.005)}$
Calculate Value of t: Calculate the value of t using a calculator. $\newline$ $t \approx \frac{\ln(3.3333\ldots)}{(12 \times \ln(1.005))}$ $\newline$ $t \approx \frac{1.0986122886681098}{(12 \times 0.0049875621120890275)}$ $\newline$ $t \approx \frac{1.0986122886681098}{0.05985074534506833}$ $\newline$ $t \approx 18.354648862953$
Convert to Years: Convert the time from months to years by dividing by $12$ . $\newline$ $t \approx 18.354648862953 / 12$ $\newline$ $t \approx 1.52955407191275$ $\newline$ Round to the nearest tenth of a year. $\newline$ $t \approx 1.5$ years