Question

what steady annual rate of inflation would cause the value of an item to rise from $2500 to $18400 in 25 years?

Answer 1

yeah i know i need to solve that, but i don't know how too :/ @ganeshie8

Answer 2

Alright, we use compound interest formula

\(A = P(1+r)^t\)

Answer 3

\(P\) = starting value = 2500
\(r\) = rate of growth
\(t\) = time = 25

Answer 4

plug them and solve \(r\) :

\(\large 18400 = 2500(1+r)^{25}\)

Answer 5

start by dividing 2500 both sides

Answer 6

how do you do that @ganeshie8

Answer 7

\(\large 18400 = 2500(1+r)^{25}\)


divide 2500 both sides, you get :
\(\large \dfrac{18400}{2500} = (1+r)^{25}\)

which is same as :
\(\large \dfrac{184}{25} = (1+r)^{25}\)

Answer 8

fine, so far ?

Answer 9

yeah, i did that @ganeshie8

Answer 10

good, next take `25th root` both sides

Answer 11

\(\large 18400 = 2500(1+r)^{25}\)


divide 2500 both sides, you get :
\(\large \dfrac{18400}{2500} = (1+r)^{25}\)

which is same as :
\(\large \dfrac{184}{25} = (1+r)^{25}\)

take 25th root both sides :
\(\large \left(\dfrac{184}{25}\right)^{\frac{1}{25}} = (1+r)\)

Answer 12

lastly subtract 1 both sides

Answer 13

\(\large 18400 = 2500(1+r)^{25}\)


divide 2500 both sides, you get :
\(\large \dfrac{18400}{2500} = (1+r)^{25}\)

which is same as :
\(\large \dfrac{184}{25} = (1+r)^{25}\)

take 25th root both sides :
\(\large \left(\dfrac{184}{25}\right)^{\frac{1}{25}} = (1+r)\)

subtract 1 both sides :
\(\large \left(\dfrac{184}{25}\right)^{\frac{1}{25}}-1 = r\)

Answer 14

plug the left chunk into ur calculator

Answer 15

http://www.wolframalpha.com/input/?i=%5Cleft%28%5Cdfrac%7B184%7D%7B25%7D%5Cright%29%5E%7B%5Cfrac%7B1%7D%7B25%7D%7D-1+

Answer 16

you get 0.083,
which is same as 8.3%

Answer 17

okay, thank you so much!! @ganeshie8

Answer 18

you're welcome !