Question

NO LINKS!! Please help me with this problem​

Answer 1

Used formula

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

$\\ \implies \sf 1200=600\left(1.0025\right)^{12t}$

$\\ \implies \sf \dfrac{1200}{600}=\left(1.0025\right)^{12t}$

$\\ \implies \sf 2=\left(1.0025\right)^{12t}$

• Apply natural logarithm on both sides

$\implies \sf \ln 2=\ln \left(1.0025\right)^{12t}$

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

$\\ \implies t=\dfrac{ \ln 2}{12 \ln (1.0025)}$

$\\ \implies t=23.1years$

Answer 2

Answer:

23.1 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 = $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:

$\implies \sf 1200=600\left(1+\dfrac{0.03}{12}\right)^{12t}$

$\implies \sf 1200=600\left(1.0025\right)^{12t}$

$\implies \sf \dfrac{1200}{600}=\left(1.0025\right)^{12t}$

$\implies \sf 2=\left(1.0025\right)^{12t}$

Take natural logs:

$\implies \sf \ln 2=\ln \left(1.0025\right)^{12t}$

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

$\implies t=\dfrac{ \ln 2}{12 \ln (1.0025)}$

$\implies t=23.13377513...$

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

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