Question

NO LINKS!! Please help me with this problem​

Answer 1

Used formula

`\boxed{\sf A=P(1+(r)/(n))^(nt)}`

`\\ \implies \sf 1200=600\left(1.0025\right)^(12t)`

`\\ \implies \sf (1200)/(600)=\left(1.0025\right)^(12t)`

`\\ \implies \sf 2=\left(1.0025\right)^(12t)`

• Apply natural logarithm on both sides

`\implies \sf \ln 2=\ln \left(1.0025\right)^(12t)`

`\\ \implies \sf \ln 2=12t\ln \left(1.0025\right)`

`\\ \implies t=( \ln 2)/(12 \ln (1.0025))`

`\\ \implies t=23.1years`

Answer 2

23.1 years (nearest tenth)

Explanation:

Compound Interest Formula

`\large \text{$ \sf A=P\left(1+(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 = $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:

`\implies \sf 1200=600\left(1+(0.03)/(12)\right)^(12t)`

`\implies \sf 1200=600\left(1.0025\right)^(12t)`

`\implies \sf (1200)/(600)=\left(1.0025\right)^(12t)`

`\implies \sf 2=\left(1.0025\right)^(12t)`

Take natural logs:

`\implies \sf \ln 2=\ln \left(1.0025\right)^(12t)`

`\implies \sf \ln 2=12t\ln \left(1.0025\right)`

`\implies t=( \ln 2)/(12 \ln (1.0025))`

`\implies t=23.13377513...`

`\implies t=23.1\: \sf years \:\:(nearest\:tenth)`

The money will double in value in approximately
`\boxed{\sf 23.1}` years.