Answer:
23.1 years (nearest tenth)
Step-by-step explanation:
Compound Interest Formula
[tex]\large \text{$ \sf A=P\left(1+\frac{r}{n}\right)^{nt} $}[/tex]
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 = $1200
- P = $600
- r = 3% = 0.03
- n = 12 (as compounded monthly)
- t = years
Substitute the given values into the formula and solve for t:
[tex]\implies \sf 1200=600\left(1+\dfrac{0.03}{12}\right)^{12t}[/tex]
[tex]\implies \sf 1200=600\left(1.0025\right)^{12t}[/tex]
[tex]\implies \sf \dfrac{1200}{600}=\left(1.0025\right)^{12t}[/tex]
[tex]\implies \sf 2=\left(1.0025\right)^{12t}[/tex]
Take natural logs:
[tex]\implies \sf \ln 2=\ln \left(1.0025\right)^{12t}[/tex]
[tex]\implies \sf \ln 2=12t\ln \left(1.0025\right)[/tex]
[tex]\implies t=\dfrac{ \ln 2}{12 \ln (1.0025)}[/tex]
[tex]\implies t=23.13377513...[/tex]
[tex]\implies t=23.1\: \sf years \:\:(nearest\:tenth)[/tex]
The money will double in value in approximately [tex]\boxed{\sf 23.1}[/tex] years.
