1. ## log

1. if a car depreciates at 15% per year, how long until it is worth half its original value?

2. led zepellin once played a concert with exerted sounds of 10 000 times the intensity of a sound that can just be heard. what is the loudness in decibels?

2. Originally Posted by william
1. if a car depreciates at 15% per year, how long until it is worth half its original value?
[snip]
Hi William,

If a Car has value 100 today, in 1 year's time it will depreciate by 15% i.e. 100 * 0.15, So:

today 100
year 1 100 - 100 * 0.15
year 2 (100 - 100 * 0.15) - (100 - 100 * 0.15)* 0.15
etc

you can see the pattern: Where A is the present value and r is the rate
A(1 - r) - A(1 - r).r = A(1- r) [1 - r] = A(1 - r)(1 - r)

So in general the depreciation is:

$A(1 - r)^{y-1}$ where y is the period in years

So:

$A(1 - r) ^{y-1} = A / 2 \: :\$(to reduce the value by one half).

$(1 - r)^{y-1} = 0.5$

$(y - 1) log (1 - r) = log(0.5)$

$y = 1 + \frac{log(0.5)}{log(0.85)}$

period = 5.26 years

3. Originally Posted by s_ingram
Hi William,

If a Car has value 100 today, in 1 year's time it will depreciate by 15% i.e. 100 * 0.15, So:

today 100
year 1 100 - 100 * 0.15
year 2 (100 - 100 * 0.15) - (100 - 100 * 0.15)* 0.15
etc

you can see the pattern: Where A is the present value and r is the rate
A(1 - r) - A(1 - r).r = A(1- r) [1 - r] = A(1 - r)(1 - r)

So in general the depreciation is:

$A(1 - r)^{y-1}$ where y is the period in years

So:

$A(1 - r) ^{y-1} = A / 2 \: :\$(to reduce the value by one half).

$(1 - r)^{y-1} = 0.5$

$(y - 1) log (1 - r) = log(0.5)$

$y = 1 + \frac{log(0.5)}{log(0.85)}$

period = 5.26 years

Why do you have y-1 as the exponent is your equation. By doing that, if you let y=1 which would be 1 year, your formula would just yield A, meaning no depreciation took place

4. Hi Artvandalay11,

your right, it should just by y not y-1. That's the problem when you try answering questions at 2.00 am! Thanks for the correction.