how long will it take for a $6000 investment to grow to $6924 at an annual rate of 9.8%, compounded…

Explanation:

Step1: Identify the 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. Given $P = 6000$, $A=6924$, $r = 0.098$ (since $9.8%=0.098$), and $n = 2$ (compounded semiannually). Substitute these values into the formula: $6924=6000(1 +\frac{0.098}{2})^{2t}$.

Step2: Simplify the equation

First, simplify the inside of the parentheses: $1+\frac{0.098}{2}=1 + 0.049=1.049$. The equation becomes $6924 = 6000(1.049)^{2t}$. Divide both sides of the equation by $6000$: $\frac{6924}{6000}=(1.049)^{2t}$. $\frac{6924}{6000}=1.154$, so $1.154=(1.049)^{2t}$.

Step3: Take the natural logarithm of both sides

$\ln(1.154)=\ln((1.049)^{2t})$. Using the property of logarithms $\ln(a^{b})=b\ln(a)$, we get $\ln(1.154)=2t\ln(1.049)$.

Step4: Solve for $t$

We know that $\ln(1.154)\approx0.143$ and $\ln(1.049)\approx0.048$. So, $0.143 = 2t\times0.048$. First, simplify the right - hand side: $2t\times0.048 = 0.096t$. Then, solve for $t$: $t=\frac{0.143}{0.096}\approx1.49$.

Answer:

$1.49$