Depreciation problem: If value dep. 20% per year, when will

[TOBYMAN]

A car was purchased in 2002 for $20,000. If its value depreciates by 20% per year, in what year was (or will be) the car worth $8192?

 

[mmm4444bot]

This web site is a cut above the usual answer mill.

The way it works here is that you show whatever work or explain whatever reasoning that you're able to accomplish so far, and then we know where to help you.

Do you understand the question; that is, do you understand what it is that you need to find?

Do you know how to calculate 20% of 20000?

Do you know? This depreciatation can be modeled using a linear function.

 

[TOBYMAN]

Re: Depreciation problem

I get this far:

.4096=(.08)^x-2002
log.4096=log (.08)^x-2002
log.4096/log.08=(x-2002)*log(.08)

 

[coloredmath]

Re: Depreciation problem

that's a weird way of finding that.

2002 - $20000
20000-(20000*.2) = 16000
2003 - $16000
16000-(16000*.2) = 12800
2004 - $12800
And so forth.

 

[mmm4444bot]

.4096=(.08)^x-2002

Okay, I guess you do know.

This depreciation is modeled using exponential growth.

Instead of putting the year 2002 into the exponent, how about letting x simply represent the number of years elapsed since 2002, instead. We can add x years to 2002 once we find it.

I hope that you intended to type 0.8^x.

v = 20000(0.8^x)

When v = 8192, then you would go through the same steps that you posted to arrive at the following.

log(0.4096) = log(0.8^x)

Can you take it from here?