Question

Consider an investment of $6000 that earns 4.5% interest.

Answer 1

Answer:

20.4 years (nearest tenth)

Step-by-step explanation:

Compound Interest Formula

$\large \sf A=P\left(1+\frac{r}{n}\right)^{nt}$

where:

• A = final amount
• P = principal amount
• r = interest rate (in decimal form)
• n = number of times interest applied per time period
• t = number of time periods elapsed

Given:

• A = $15,000
• P = $6,000
• r = 4.5% = 0.045
• n = 12 (monthly)

Substitute the given values into the formula and solve for t:

$\implies \sf 15000=6000\left(1+\frac{0.045}{12}\right)^{12t}$

$\implies \sf \dfrac{15000}{6000}=\left(1.00375\right)^{12t}$

$\implies \sf 2.5=\left(1.00375\right)^{12t}$

$\implies \sf \ln (2.5)=\ln \left(1.00375\right)^{12t}$

$\implies \sf \ln (2.5)=12t \ln \left(1.00375\right)$

$\implies \sf t=\dfrac{\ln (2.5)}{12 \ln (1.00375)}$

$\implies \sf t=20.40017123$

Therefore, it would take 20.4 years (nearest tenth) for the investment to reach $15,000.