how long does it take to double a $1,000 investment that pays 6.5% annual interest, compounded monthly…

how long does it take to double a $1,000 investment that pays 6.5% annual interest, compounded monthly? which equation can you use to solve this problem? 1000=(1 + \\frac{0.065}{12})^t 2=(1 + \\frac{0.065}{12})^{12t} 2=(1 + \\frac{6.5}{12})^{12t} done

how long does it take to double a $1,000 investment that pays 6.5% annual interest, compounded monthly? which equation can you use to solve this problem? 1000=(1 + \\frac{0.065}{12})^t 2=(1 + \\frac{0.065}{12})^{12t} 2=(1 + \\frac{6.5}{12})^{12t} done

Answer

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 form), $n$ is the number of times interest is compounded per year, and $t$ is the number of years. We want to double the investment, so $A = 2P$. Given $P=$1000$, $r = 0.065$ (since $6.5%=0.065$), and $n = 12$ (compounded monthly). Substituting these values into the formula: $2P=P(1 +\frac{0.065}{12})^{12t}$. Since $P\neq0$, we can divide both sides of the equation by $P$.

Step2: Simplify the equation

Dividing both sides of $2P=P(1+\frac{0.065}{12})^{12t}$ by $P$ gives $2=(1+\frac{0.065}{12})^{12t}$.

Answer:

$2=(1+\frac{0.065}{12})^{12t}$