how many years will it take for a $400 investment to grow to $1000 with a interest rate of 12% compounded…

Explanation:

Step1: Recall compound - interest formula

The compound - interest formula is $A = P(1+\frac{r}{n})^{nt}$, where $A$ is the final amount, $P$ is the principal amount, $r$ is the annual interest rate (in decimal), $n$ is the number of times interest is compounded per year, and $t$ is the number of years. Here, $P = 1000$, $A=4000$, $r = 0.12$ (since $12%=0.12$), and $n = 12$ (compounded monthly). So, $4000=1000(1 +\frac{0.12}{12})^{12t}$.

Step2: Simplify the equation

First, divide both sides of the equation by $1000$: $\frac{4000}{1000}=(1 + 0.01)^{12t}$, which simplifies to $4=(1.01)^{12t}$.

Step3: Take the natural logarithm of both sides

$\ln(4)=\ln((1.01)^{12t})$. Using the property of logarithms $\ln(a^b)=b\ln(a)$, we get $\ln(4)=12t\ln(1.01)$.

Step4: Solve for $t$

$t=\frac{\ln(4)}{12\ln(1.01)}$. We know that $\ln(4)\approx1.3863$ and $\ln(1.01)\approx0.00995$. $t=\frac{1.3863}{12\times0.00995}=\frac{1.3863}{0.1194}\approx11.61$.

Answer:

Approximately $11.61$ years.