Question

PLEASE PLEASE HELP
The consumer price index (CPI) indicates the relative change in price over time for a fixed basket of goods and service. In general if the rat of inflation averages r% over n years, then the CPI after n years is CPI=CPI (base 0) (1+(r/100))^n, where CPI(base 0) is the CPI index at the beginning of the n-years.

(a) the CPI was 151.5 for 1995 and 202.4 for 2003. Assuming that annual inflation remained constant for this time period, determine the average annual inflation rate.
a) the average annual inflation rate from 1995 to 2003 was approximately__%

(b) Using the inflation rate from part a in what year will the CPI reach 256?
b) the CPI will reach 256 in the year ________

Answer 1

\[C=C_0\left(1+{r\over100}\right)^n\]

Answer 2

a) C0=151.5
C=202.4 , n = 2003-1995=8
find "r"

Answer 3

so it will look like\[202.4=151.5\left( 1+\frac{ r }{ 100 } \right)^{2003-1995}\]

Answer 4

yep

Answer 5

I got a negative number its -83.3 which i think is wrong?

Answer 6

how do you get r by itself since i think I am doing wrong?

Answer 7

first divide both sides by 151.5
then take logarithms to bring the exponents down

Answer 8

then simplify
then take inverse logarithm to get the "r" out

Answer 9

can you refresh me on how to take the logarithms to bring the exponent

Answer 10

enter the simplified form first.

Answer 11

okay

Answer 12

so i did 202.4/151.5

Answer 13

k..

Answer 14

so how will you get rid of the exponent?

Answer 15

by using log

Answer 16

\[\Large{\log(a^m)=m\times\log(a)}\]

Answer 17

so it will be log(202.4/151.5)=8log(1+(r/100))

Answer 18

exactly

Answer 19

simplify it to \[\log(\ldots)=\ldots\]

Answer 20

okay so would it look like log202.4)-log(151.5)=8log(1+(r/100)) and thats all I could think of

Answer 21

you can use a calc to get the numbers...

Answer 22

but I will you put them all to one side and r by itself? from that

Answer 23

yes. you'd get a simple form like I showed above..

Answer 24

log of 1+r/100 on one side and a number on the other side..

Answer 25

so it will look like log of 1 (r/100) =.125797

Answer 26

@electrokid

Answer 27

good.
now, I presume that is "log base 10"
then
\[1+{r\over100}=10^{\rm number}\]

Answer 28

10^number you got on the right side

Answer 29

okay?

Answer 30

can you show me how it will look like

Answer 31

just did..
look two steps above. :)

Answer 32

remove the "log" on the left,
and raise "10" to the power of that number

Answer 33

so it will look like 1+(r/100)=10^.25797

Answer 34

exactly..
solve for "r" like a good old algebra problem.

Answer 35

i got 33.60%

Answer 36

\[{1\over8}\log\left(202.4\over151.5\right)=\log\left(1+{r\over100}\right)\]

Answer 37

re do the left part

Answer 38

APR values are usually < 10%

Answer 39

i got 3.69

Answer 40

how will you do part b?

Answer 41

good.

Answer 42

use this "r"
the C0 = C for 1998
C = 256,
find "n"

Answer 43

similar steps.. no need to take 10^
just the log will do

the "n" is the number of years after 1998.
so, find what year it'd be

Answer 44

i keep getting a neg. number

Answer 45

@electrokid

Answer 46

follow the same steps
should get it

Answer 47

\[256=151.5\left(1+{3.7\over100}\right)^n\]

Answer 48

okay so i did log(256)-log151)=nlog(1+.037)

Answer 49

how will you get n buy itself from that? @electrokid

Answer 50

is there a chance you can show me how it will look like if n was by itself?

Answer 51

\[\log\left(256\over151\right)=n\times\log(1+0.037)\]

Answer 52

yes now how do you just get n by itself. so just divided?

Answer 53

yes..
use a calculator to gt the logs and divide.

Answer 54

i got 70

Answer 55

but wouold the 256 be 202.4 instead and 252 be 151.5

Answer 56

no.
it'd be 151.5
the answer is n=14.5 (approx)

Answer 57

then where did the 256 come from again?

Answer 58

check with the quesrtion

Answer 59

oh okay thank you! sorry was confused!!! thank you for helping me while i struggle!

Answer 60

:)
you are welcome.