# Exponential Growth and Decay (word problems).

• Apr 3rd 2007, 08:24 PM
Mulya66
Exponential Growth and Decay (word problems).
1) "Solve the equation: The price of a new car is \$28,000 and it depreciates 9% each year. How much is the car worth in 6 years?"

I THINK #1 is as follows...

y = 28,000(1+/-0.09)^6
y = 28,000(+/-1.09)^6
y = 28,000(1.678)
y = 46,984
And this looks completely wrong to me.

2) "Solve the equation: A population of 175 snails is increasing at an annual rate of 12%. At this rate, how long will it take for the population of snails to reach 315?"

I have absolutely no idea.

I know that I should apply the formulas y = ae^+/-kt and y = a(1+/-r)^t, but I'm not quite sure for which type of problem uses which formula. I do believe that y = a(1+/-r)^t is used for money, but again, I'm just stuck on how to do these. Thanks in advance to anyone who'll tackle these.
• Apr 3rd 2007, 08:36 PM
CaptainBlack
Quote:

Originally Posted by Mulya66
1) "Solve the equation: The price of a new car is \$28,000 and it depreciates 9% each year. How much is the car worth in 6 years?"

I THINK #1 is as follows...

y = 28,000(1+/-0.09)^6

The car depreciates at 9%, so the rate of growth of value is -0.09 so in six years the car is worth:

y = 28000(1 - 0.09)^6 = 28000 0.91^6 ~= 15900.34

Quote:

y = 28,000(+/-1.09)^6
y = 28,000(1.678)
y = 46,984
1+/-0.09 is 0.91 or 1.09, not +/-1.09

RonL
• Apr 3rd 2007, 08:42 PM
CaptainBlack
Quote:

Originally Posted by Mulya66
1) 2) "Solve the equation: A population of 175 snails is increasing at an annual rate of 12%. At this rate, how long will it take for the population of snails to reach 315?"

After N years the population is:

P(N) = 175 (1+0.12)^N.

If the population after N years is 315 then we hav:

315 = 175 (1.12)^N

which we have to solve.

rearranging this gives:

(1.12)^N = 315/175 = 1.8

Now take logs:

N log(1.12) = log(1.8),

so:

N = log(1.8)/log(1.12) ~= 5.19 years.

RonL
• Apr 3rd 2007, 08:45 PM
Mulya66
Oh okay, now I understand my error in #1.

Thank you!