find the time required for an investment of $10,000 to grow to $19,000 at an annual interest rate of 5%…

find the time required for an investment of $10,000 to grow to $19,000 at an annual interest rate of 5%. complete parts a. through e.\na. find the time required for an investment of $10,000 to grow to $19,000 at an annual interest rate of 5% if the interest is compounded yearly. 13.155 years (simplify your answer. do not round until the final answer. then round to the nearest thousandth as needed.)\nb. find the time required for an investment of $10,000 to grow to $19,000 at an annual interest rate of 5% if the interest is compounded quarterly. 12.917 years (simplify your answer. do not round until the final answer. then round to the nearest thousandth as needed.)\nc. find the time required for an investment of $10,000 to grow to $19,000 at an annual interest rate of 5% if the interest is compounded monthly. 12.864 years (simplify your answer. do not round until the final answer. then round to the nearest thousandth as needed.)\nd. find the time required for an investment of $10,000 to grow to $19,000 at an annual interest rate of 5% if the interest is compounded daily. □ years (simplify your answer. do not round until the final answer. then round to the nearest thousandth as needed.)

find the time required for an investment of $10,000 to grow to $19,000 at an annual interest rate of 5%. complete parts a. through e.\na. find the time required for an investment of $10,000 to grow to $19,000 at an annual interest rate of 5% if the interest is compounded yearly. 13.155 years (simplify your answer. do not round until the final answer. then round to the nearest thousandth as needed.)\nb. find the time required for an investment of $10,000 to grow to $19,000 at an annual interest rate of 5% if the interest is compounded quarterly. 12.917 years (simplify your answer. do not round until the final answer. then round to the nearest thousandth as needed.)\nc. find the time required for an investment of $10,000 to grow to $19,000 at an annual interest rate of 5% if the interest is compounded monthly. 12.864 years (simplify your answer. do not round until the final answer. then round to the nearest thousandth as needed.)\nd. find the time required for an investment of $10,000 to grow to $19,000 at an annual interest rate of 5% if the interest is compounded daily. □ years (simplify your answer. do not round until the final answer. then round to the nearest thousandth as needed.)

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 are given $P = 10000$, $A=19000$, $r = 0.05$. When compounded daily, $n = 365$. We need to solve the formula for $t$: [ \begin{align*} A&=P(1 +\frac{r}{n})^{nt}\ \frac{A}{P}&=(1+\frac{r}{n})^{nt}\ \ln(\frac{A}{P})&=nt\ln(1 +\frac{r}{n})\ t&=\frac{\ln(\frac{A}{P})}{n\ln(1+\frac{r}{n})} \end{align*} ]

Step2: Substitute values

Substitute $A = 19000$, $P = 10000$, $r = 0.05$, and $n = 365$ into the formula for $t$: [ \begin{align*} t&=\frac{\ln(\frac{19000}{10000})}{365\ln(1+\frac{0.05}{365})}\ &=\frac{\ln(1.9)}{365\ln(1+\frac{0.05}{365})}\ \end{align*} ] First, calculate $1+\frac{0.05}{365}=1+\frac{1}{7300}=\frac{7301}{7300}\approx1.000137$. Then $\ln(1+\frac{0.05}{365})\approx\ln(1.000137)\approx0.000137$. And $\ln(1.9)\approx0.641854$. [ \begin{align*} t&=\frac{0.641854}{365\times0.000137}\ &=\frac{0.641854}{0.050005}\ &\approx12.836 \end{align*} ]

Answer:

$12.836$