if you borrow $4700 at an apr of r (as a decimal) from a lending institution that compounds interest…

if you borrow $4700 at an apr of r (as a decimal) from a lending institution that compounds interest continuously, and if you wish to pay off the note in 3 years, then your monthly payment m, in dollars, can be calculated using $m = \\frac{4700(e^{r/12}-1)}{1 - e^{-3r}}$. your budget will allow a payment of $150 per month, and you are shopping for an interest rate that will give a payment of this size. what interest rate do you need to find? (round your answer to two decimal places.)

if you borrow $4700 at an apr of r (as a decimal) from a lending institution that compounds interest continuously, and if you wish to pay off the note in 3 years, then your monthly payment m, in dollars, can be calculated using $m = \\frac{4700(e^{r/12}-1)}{1 - e^{-3r}}$. your budget will allow a payment of $150 per month, and you are shopping for an interest rate that will give a payment of this size. what interest rate do you need to find? (round your answer to two decimal places.)

Answer

Explanation:

Step1: Set up the equation

We are given $M = 150$ and the formula $M=\frac{4700(e^{r/12}-1)}{1 - e^{-3r}}$. So, $150=\frac{4700(e^{r/12}-1)}{1 - e^{-3r}}$.

Step2: Cross - multiply

$150(1 - e^{-3r})=4700(e^{r/12}-1)$. $150-150e^{-3r}=4700e^{r/12}-4700$.

Step3: Rearrange the terms

$4700e^{r/12}+150e^{-3r}=4700 + 150=4850$. Let $x = e^{r/12}$, then $e^{-3r}=\frac{1}{e^{3r}}=\frac{1}{(e^{r/12})^{36}}=\frac{1}{x^{36}}$. The equation becomes $4700x+\frac{150}{x^{36}}=4850$. This is a non - linear equation. We can use a numerical method like Newton - Raphson method or use a graphing utility. Using a graphing utility or a calculator with equation - solving capabilities: We can rewrite the equation as $4700x^{37}-4850x^{36}+150 = 0$. Solving for $x$ and then for $r$ (since $x = e^{r/12}$, so $r = 12\ln(x)$). After using a calculator or software to solve the equation, we find $r\approx0.07$.

Answer:

$7.00$