math challenge!!!!!?

Question

i need help with the right formula and which numbers go where; thanks so much if you can help!!

Inflation, as measured by the U.S. consumer price index, increased by 2.8% in the year 2001. If this rate were to continue for the next 10 years, determine how long before the value of a dollar would be reduced to 90 cents.

p.s. i need to use either log or ln to solve this problem, thanks!

Answer 1

OK... one dollar today is 1.00
One dollar a year from now is 1.00 * (1-0.028)
One dollar two years from now is 1,00 * (1-0.028)^2
One dollar n years from now is 1.00 * (1-0.028)^n

The question is: 0.9 = 1 * (1-0.028)^x, solve for x.

Since your variable is in an exponent, you take logs.
ln 0.9 = ln (0.972)^x

By one of the log rules, you can haul the exponent in front of the log:
ln 0.9 = x * ln(0.972)

Then divide to get:
x = (ln 0.9)/(ln 0.972)
x = 3.7

The answer is 3.7 years.

Answer 2

The price of an item that cost $1.00 in 2000 is $1.028 in 2001. It's 2001 value, then, would be $1*(1/1.028)
let
F= Final value of an item =$0.90
P = initial value of the item in the year 2000 = $1.00
i = inflation rate = 0.028
F = P*(1/1+i)^n
ln(F/P) = n*ln(1/1+i)
n = (ln(F/P))/ln(1/1+i) in years
or
n = (ln(P/F))/ln(1+i)
You can use log instead of ln:
n = (log(P/F))/log(1+i)

Answer 3

v(t)=e^rt, where v(t) denotes the value of the dollar as a function of time t, r=rate at which the dollar inflates, and t=time (years).

So by plugging in for known constants, we obtain:

0.9 = exp [ -0.028 t ] OR 0.9 = e^(-0.028t)

Solving for t, we get

ln ( 9/10 ) = -0.028 t

t = ln ( 9/10 ) / -0.028

which is approximately

3.762875559 years, OR

3 years, 278 days, 10 hours, 47 minutes, 23.635 seconds

Answer 4

90=100(1-0.028)^n
(0.972)^n=10/9
nlog 0.972=log10-log9
n log 0.972=log10-log9
=>1-0.9542=0.0458
n=4.6 years

Answer 5

How about..........you ask a math teacher!

DUHHHH.