How do you evaluate 1200=300(1+r)^{5}1200=300(1+r)5?
2 Answers
Explanation:
1200=300(1+r)^51200=300(1+r)5
First, divide both sides of the equation by
1200/300=(300(1+r)^5)/3001200300=300(1+r)5300
4=(1+r)^54=(1+r)5
To undo the power of
4^(1/5)=((1+r)^5)^(1/5)415=((1+r)5)15
4^(1/5)=1+r415=1+r
Finally, subtract
r=4^(1/5)-1r=415−1
Personally, I prefer the simplification
r=2^(2/5)-1r=225−1
Explanation:
We have:
1200=300(1+r)^5 1200=300(1+r)5
Let
1200 = 300z^5 => z^5 = 4 1200=300z5⇒z5=4
First, we will put the equation into polar form:
|omega| = 4|ω|=4
theta =0θ=0
So then in polar form we have:
z^5 = 4(cos0 + isin0) z5=4(cos0+isin0)
We now want to solve the equation
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
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
After which the pattern continues (due the above mentioned periodicity).
We can plot these solutions on the Argand Diagram:
