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

2 Answers
mason m
Aug 31, 2017

#r=2^(2/5)-1#

Explanation:

#1200=300(1+r)^5#

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

#1200/300=(300(1+r)^5)/300#

#4=(1+r)^5#

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

#4^(1/5)=((1+r)^5)^(1/5)#

#4^(1/5)=1+r#

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

#r=4^(1/5)-1#

Personally, I prefer the simplification #4^(1/5)=(2^2)^(1/5)=2^(2/5)#:

#r=2^(2/5)-1#

Steve M
Sep 3, 2017

Stevem

Explanation:

We have:

# 1200=300(1+r)^5 #

Let # omega=1+r #, and let #z^5 = omega#, then:

# 1200 = 300z^5 => 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(cos0 + isin0) #

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 #2pi#, so we can equivalently write (incorporating the periodicity):

# z^5 = 4(cos(0+2npi) + isin(0+2npi)) \ \ \ n in ZZ #

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

# z = (4(cos(0+2npi) + isin(0+2npi)))^(1/5) #
# \ \ = 4^(1/5)(cos(2npi) + isin(2npi)))^(1/5) #
# \ \ = 4^(1/5)(cos((2npi)/5) + isin((2npi)/5)) #
# \ \ = 4^(1/5)(costheta + isintheta) #

Where:

# theta =(2npi)/5#

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

Stevem

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

We can plot these solutions on the Argand Diagram:

Wolfram Alpha

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