# How do you evaluate 1200=300(1+r)^{5}?

Aug 31, 2017

$r = {2}^{\frac{2}{5}} - 1$

#### Explanation:

$1200 = 300 {\left(1 + r\right)}^{5}$

First, divide both sides of the equation by $300$:

$\frac{1200}{300} = \frac{300 {\left(1 + r\right)}^{5}}{300}$

$4 = {\left(1 + r\right)}^{5}$

To undo the power of $5$, raise both sides of the equation to the $1 / 5$ power:

${4}^{\frac{1}{5}} = {\left({\left(1 + r\right)}^{5}\right)}^{\frac{1}{5}}$

${4}^{\frac{1}{5}} = 1 + r$

Finally, subtract $1$ from both sides of the equation:

$r = {4}^{\frac{1}{5}} - 1$

Personally, I prefer the simplification ${4}^{\frac{1}{5}} = {\left({2}^{2}\right)}^{\frac{1}{5}} = {2}^{\frac{2}{5}}$:

$r = {2}^{\frac{2}{5}} - 1$

Sep 3, 2017

#### Explanation:

We have:

$1200 = 300 {\left(1 + r\right)}^{5}$

Let $\omega = 1 + r$, and let ${z}^{5} = \omega$, then:

$1200 = 300 {z}^{5} \implies {z}^{5} = 4$

First, we will put the equation into polar form:

$| \omega | = 4$
$\theta = 0$

So then in polar form we have:

${z}^{5} = 4 \left(\cos 0 + i \sin 0\right)$

We now want to solve the equation ${z}^{5} = 4$ for $z$ (to gain $5$ solutions):

Whenever dealing with complex variable equation such as this it is essential to remember that the complex exponential (and therefore the polar representation) has a period of $2 \pi$, so we can equivalently write (incorporating the periodicity):

${z}^{5} = 4 \left(\cos \left(0 + 2 n \pi\right) + i \sin \left(0 + 2 n \pi\right)\right) \setminus \setminus \setminus n \in \mathbb{Z}$

By De Moivre's Theorem we can write this as:

$z = {\left(4 \left(\cos \left(0 + 2 n \pi\right) + i \sin \left(0 + 2 n \pi\right)\right)\right)}^{\frac{1}{5}}$
 \ \ = 4^(1/5)(cos(2npi) + isin(2npi)))^(1/5)
$\setminus \setminus = {4}^{\frac{1}{5}} \left(\cos \left(\frac{2 n \pi}{5}\right) + i \sin \left(\frac{2 n \pi}{5}\right)\right)$
$\setminus \setminus = {4}^{\frac{1}{5}} \left(\cos \theta + i \sin \theta\right)$

Where:

$\theta = \frac{2 n \pi}{5}$

And we will get $5$ unique solutions by choosing appropriate values of $n$. Working to 3dp, and using excel to assist, we get:

After which the pattern continues (due the above mentioned periodicity).

We can plot these solutions on the Argand Diagram: