Question

A = P(1+i)^n

1. Isolate the variable P
2. Isolate the variable i

Answer 1

where did you get 10 from?

Answer 2

this is so confusing :(

Answer 3

Polar form...?

Answer 4

take natural log of both sides
\[\ln(A)=\ln(P(1+i)^n)=\ln(P)+\ln(1+i)^n=\ln(P)+n*\ln(1+i)\]
so we have
\[\ln(A)=\ln(P)+n*\ln(1+i)\]
subract ln(P) on both sides
\[\ln(A)-\ln(P)=n*\ln(1+i)\]
remember ln(x/y)=lnx-lny

so we can write
\[\ln(\frac{A}{P})=n*\ln(1+i)\]
now multiply both sides by 1/n

\[\frac{1}{n}*\ln(\frac{A}{P})=\frac{1}{n}*n*\ln(1+i)\]

\[\frac{1}{n}\ln(\frac{A}{P})=\ln(1+i)\]
\[\ln(\frac{A}{P})^\frac{1}{n}=\ln(1+i)\]

do e^ on both sides

\[(\frac{A}{P})^\frac{1}{n}=1+i\]

subtract i on both sides

\[(\frac{A}{P})^\frac{1}{n}-1=i\]

but this way is probably way more confusing (and unnecessary) lol

so we have
\[A=P(1+i)^n\]

we want to solve for i
so first divide P on both sides

\[\frac{A}{P}=(1+i)^n\]

now do ^(1/n) on both sides


\[(\frac{A}{P})^\frac{1}{n}=(1+i)^{n*(\frac{1}{n})}\]

\[(\frac{A}{P})^\frac{1}{n}=(1+i)^1\]

\[(\frac{A}{P})^\frac{1}{n}=1+i\]

now subtract 1 on bboth sides

\[(\frac{A}{P})^\frac{1}{n}-1=i\]

Answer 5

you probably don't really want to look at the first way i did
but the second way she be pretty easy to understand
what do you think?

Answer 6

i think the way you did it is confusing, because here on the question theyre just asking to isolate the variable for P so wouldnt i do something along the lines of P=(a)(1+i)^n

Answer 7

the second way?

Answer 8

yes

Answer 9

you want to solve for i
we need to work on getting i by itself
so we notice i is inside here: (1+i)^n right?

but (1+i)^n is being multiplied by that P

so to get to (1+i)^n by itself we need to divide both sides by P first

Answer 10

\[\frac{A}{P}=(1+i)^n\]
so you understand this part right?

Answer 11

yes

Answer 12

ok but i is inside that paranthese that is to the n power

to get rid of the n power we need to raise both sides by 1/n

becasue we know n*1/n will give us 1 right?

\[(\frac{A}{P})^\frac{1}{n}=((1+i)^{n})^\frac{1}{n}\]
do you understand this part?
do you undestand why I raised both sides to the 1/n power?

Answer 13

remember im trying to isolate the variable by themselves so im trying to isolate i by itself and p by its self using the formula A = P( 1+ i )^n

Answer 14

since n*1/n=1 we have
\[(\frac{A}{P})^\frac{1}{n}=(1+i)^1\]

Answer 15

but \[(1+i)^1=(1+i)=1+i\]

Answer 16

so we have
\[(\frac{A}{P})^\frac{1}{n}=1+i\]

to isolate i by itself now we need to subtract 1 on both sides

Answer 17

\[(\frac{A}{P})^\frac{1}{n}-1=i\]

Answer 18

if you don't like the way this is written you can always write it going the other way:
\[i=(\frac{A}{P})^\frac{1}{n}-1\]

Answer 19

they mean the samethings

Answer 20

It also works out if u do 1+i as sqrt(2) and pi/4 plus complex powers.

Answer 21

But your way is simpler...(much)

Answer 22

lol
wedphill do you understand now?

Answer 23

yes i think i get it now

Answer 24

thank you so much for your help :)

Answer 25

you aren't just saying that to get rid of me right?

Answer 26

nooo

Answer 27

lol
ok im glad i can help

Answer 28

I was just confused at the part when i had to isolate i

Answer 29

i didnt understand how to re-arrange the equation

Answer 30

your 2nd problem had a bit more steps than 1st problem
in the problem we had to undo alot of stuff to get to i

Answer 31

thank you!

Answer 32

first the i was being added to the 1
and then that was to the nth power
but even all of that was being multiplied by that P
so we would from outside to inside to get that i
we had to undo the multiplication of P by dividing by P on both side
we had to get rid of the nth power by raising both sides to the 1/n th power
and then finally to get rid of the adding 1 we had to subtract 1 on both sides

Answer 33

your welcome

Answer 34

so basically to solve for P i use P = A / (1 + i)^n
and to solve for i, i use i = (A/P)^(1/n) - 1

Answer 35

well yeah you already solved for P and i
i think you are asking me to confirm your answers
and yes that is right those are your answers

Answer 36

THANKS :)

Answer 37

i understand it now :)