Question

instructions timed test this test has a time limit of 1 hour and 40 minutes. this test will save and submit automatically when the time expires. warnings appear when half the time, 5 minutes, 1 minute, and 30 seconds remain. multiple attempts this test allows multiple attempts. force completion this test can be saved and resumed at any point until time has expired. the timer will continue to run if you leave the test. your answers are saved automatically. remaining time: 1 hour, 31 minutes, 29 seconds. question completion status: moving to another question will save this response. question 5 3 points a lump sum of $100 was deposited into an account earning 12.0% annual interest, compounded quarterly. how long, to the nearest tenth of a year, will it take for this to reach $2,000,000? enter your answer as a numerical value (no label) and round to the nearest tenth of a year.

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 $P = 100$, $A=2000000$, $r = 0.12$ (since $12\%=0.12$), and $n = 4$ (compounded quarterly).
Substitute these values into the formula: $2000000=100(1 +\frac{0.12}{4})^{4t}$.

Step2: Simplify the equation

First, divide both sides of the equation by 100: $\frac{2000000}{100}=(1 + 0.03)^{4t}$, which simplifies to $20000 = 1.03^{4t}$.

Step3: Take the natural logarithm of both sides

$\ln(20000)=\ln(1.03^{4t})$.
Using the property of logarithms $\ln(a^{b})=b\ln(a)$, we get $\ln(20000)=4t\ln(1.03)$.

Step4: Solve for $t$

First, find $\ln(20000)\approx9.9035$ and $\ln(1.03)\approx0.0296$.
Then, $t=\frac{\ln(20000)}{4\ln(1.03)}$.
Substitute the values of the logarithms: $t=\frac{9.9035}{4\times0.0296}=\frac{9.9035}{0.1184}\approx83.6$.

Answer:

83.6