math

slothz:

A machine purchased for $400 000 depreciates at a rate of 20% p.a. After how many years
will the machine have a value of $131 072?
Just don't really know the formula for calculating n plz help
Cheers guys

hI,
what nice manners , thank you.

It is almost the same formula as compound interest which is S=P(1+r) ⁿ
Where S is the future value,
p is the present (or initial) value,
r is the interest rate per compounding period
n is the number of compounding periods

The only difference is that if interest is compounding the amount is getting bigger so it is (1+r)
If the item is depreciating the amount is getting smaller so you use (1-r)

A machine purchased for $400 000 depreciates at a rate of 20% p.a. After how many years will the machine have a value of $131 072?
So the formula you need to use is S=P(1 -r) ⁿ
S is the future value = $131 072
P is the initial value = $400 000
r = 0.2
n = unknown

131 072 = 400 000(1-0.2) ⁿ
divide both sides by 400 000
0.32768 = (0.8) ⁿ

Now, whenever the exponent is the unknown, you have to use logs to solve the question
It doesn't matter what base you use, the calculator deals only with 10 and e so you'd best choose one of those.

log 0.32768 = log (0.8) ⁿ

log 0.32768 = n * log 0.8

n = log 0.32768 / log 0.8
and you can finish it.

Here are the log identities, there are a lot that you are expected to remember.
http://en.wikipedia.org/wiki/List_of_logarithmic_identities