application problem: depreciation of photocopier

[jshaziza]

A color photocopier is purchased for $5200. Its value each year is about 80% of its value in the preceding year. Its value in dollars after t years is given by the exponential function V(t)=5200(0.8)^t

a. After what amount of time will the salvage value be $1200?

What would the formula be to solve for t? Thx. for any help.

 

[Denis]

Re: application problem

.8^t = 1200/5200

if a^x = b, then x = log(b) / log(a)

 

[jshaziza]

Using your formula I got 0.8286. How would this translate to amount of time for the salvage value to be $1200? And what is principle or name of the formula you gave me, a^x=b, then x=log(b)/log(a), I coudn't find it the section I am working in. Thx. for your help.

 

[jwpaine]

Using your formula I got 0.8286. How would this translate to amount of time for the salvage value to be $1200? And what is principle or name of the formula you gave me, a^x=b, then x=log(b)/log(a), I coudn't find it the section I am working in. Thx. for your help.

http://www.purplemath.com/modules/logs.htm
http://www.purplemath.com/modules/logrules4.htm << change of base formula

 

[Denis]

Using your formula I got 0.8286.

log(1200/5200) / log(.8) = 6.5712 : HOW did you ever come up with .8286 :shock:

 

[jshaziza]

Denis, my mistake, I forgot to take the log of a, but would I put as my answer 6.6 years?

 

[Denis]

How would we know; ask your teacher what he/she expects as rounding off.