if tanisha has $1000 to invest at 6% per annum compounded monthly, how long will it be before she has $1600…

Explanation:

Step1: Recall compound - interest formula for monthly compounding

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 compounded per year, and $t$ is the number of years. Given $P = 1000$, $A=1600$, $r = 0.06$ (since $6%=0.06$), and $n = 12$ (compounded monthly). So, $1600=1000(1 +\frac{0.06}{12})^{12t}$.

Step2: Simplify the equation

First, divide both sides of the equation by 1000: $\frac{1600}{1000}=(1 + 0.005)^{12t}$, which simplifies to $1.6=(1.005)^{12t}$.

Step3: Take the natural logarithm of both sides

$\ln(1.6)=\ln((1.005)^{12t})$. Using the property of logarithms $\ln(a^b)=b\ln(a)$, we get $\ln(1.6)=12t\ln(1.005)$.

Step4: Solve for $t$

$t=\frac{\ln(1.6)}{12\ln(1.005)}$. Calculate $\ln(1.6)\approx0.4700036292$ and $\ln(1.005)\approx0.0049875427$. Then $12\ln(1.005)\approx0.0598505124$. So, $t=\frac{0.4700036292}{0.0598505124}\approx7.8530\approx7.85$ (rounded to two decimal places).

For continuous compounding:

Step1: Recall continuous - compounding formula

The formula is $A = Pe^{rt}$, where $A = 1600$, $P = 1000$, and $r=0.06$. So, $1600 = 1000e^{0.06t}$.

Step2: Simplify the equation

Divide both sides by 1000: $1.6=e^{0.06t}$.

Step3: Take the natural logarithm of both sides

$\ln(1.6)=\ln(e^{0.06t})$. Since $\ln(e^x)=x$, we have $\ln(1.6)=0.06t$.

Step4: Solve for $t$

$t=\frac{\ln(1.6)}{0.06}$. Since $\ln(1.6)\approx0.4700036292$, then $t=\frac{0.4700036292}{0.06}\approx7.83$ (rounded to two decimal places).

Answer:

Compounding monthly, it will be about $7.85$ years. Compounding continuously, it will be about $7.83$ years.