The value of a $20,000 car decreases every year, t. The equation below model this situation.

20,000(1-0.04t) = 13,000

How many years will the car be worth $13,000? Round to the nearest tenth if necessary.

I got 20,000(.96t)= 13,000, but I don't believe gives me the correct answer. Help!

Damon answered this for you yesterday, and pointed out your typo

http://www.jiskha.com/display.cgi?id=1335108312

I also agree that your equation should read

20000(1-.04)^t = 13000

then .96^t = .65
take log of both sides

log (.96^t) = log .65
use log rules ...

t(log .96) = log .65
t = log .65 / log .96 = 10.5527 = appr 10.6 years

To solve the equation, we need to isolate the variable t, which represents the number of years. Let's go through the steps together.

Starting with the equation you got: 20,000(.96t) = 13,000.

We want to isolate t on one side of the equation, so divide both sides by 20,000:

(.96t) = 13,000 / 20,000.

Simplifying the right side of the equation gives:

(.96t) = 0.65.

To isolate t, we need to divide both sides of the equation by 0.96:

(.96t) / 0.96 = 0.65 / 0.96.

Simplifying further, we get:

t = 0.65 / 0.96.

Calculating the right side of the equation gives:

t ≈ 0.677.

To round to the nearest tenth, we keep one digit after the decimal point:

t ≈ 0.7.

Therefore, the car will be worth $13,000 after approximately 0.7 years, rounded to the nearest tenth.