Exp depreciation

[johnjones]

Exponential depreciation of a car's value resulted in a 50% decrease over 5 years.
If your investment appreciates in value by 10% each year, how long will it take for a 100% increase (give years and months to nearest month).

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I've figured out the percentage decrease each year is approximately 13%.
I went y = yoa^t [yo = y not]
Decay factor: a^5 = 1/2
a = fifthroot of (1/2)
a = 0.87055
1 - 0.87055

So if it appreciates in 10%, do I change 50% to 60%? I'm kind of confused, thanks. If my method is incorrect so far, please let me know.

 

[tkhunny]

The timing of the appreciation is not clear t me. Is it simultaneous with the depreciation

(0.5)^x/5*(1.1)^x/5 for x ≤ 5
(1.1)^(x-5)/5 for x >5

or subsequent to it?

(0.5)^x/5 for x ≤ 5
(1.1)^(x-5)/5 for x >5

or are we looking at two confused problem statements?

 

[johnjones]

The timing of the appreciation is not clear t me. Is it simultaneous with the depreciation

(0.5)^x/5*(1.1)^x/5 for x ≤ 5
(1.1)^(x-5)/5 for x >5

or subsequent to it?

(0.5)^x/5 for x ≤ 5
(1.1)^(x-5)/5 for x >5

or are we looking at two confused problem statements?

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The question was given to me in two parts: (I have no clue if they are separate questions)

(a) Exponential depreciation of a car's value resulted in 50% decrease over 5 years. What was the percentage decrease each year?

(b) If your investment appreciates in value by 10% each year, how long will it take for a 100% increase?

I think I solved part a right earlier... not sure. As for part b, hmmm
:shock:

Edited by tkhunny to remove formatting errors.

 

[tkhunny]

a = annual percentage decrease
1-a = annual value factor
(1-a)⁵ = 1/2 ==> a = 12.9449%
This seems to agree with your result.
You can use logarithms to solve it.

b = annual appreciation
1+b = annual value factor
n = number of years to double (increase 100%)
(1+b)ⁿ
b = 0.10
(1+0.10)ⁿ = 2 ==> n = 7.27254 years
You need logarithms for this one.