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3 years ago

The value of a certain car decreases by 16% each year. What is the 1⁄2-life of the car?

Mathematics

1 answer:

svet-max [94.6K]3 years ago

7 0

Answer:

The half life of the car is 3.98 years.

Step-by-step explanation:

The value of the car after t years is given by the following equation:

$V(t) = V(0)(1-r)^{t}$

In which V(0) is the initial value and r is the constant decay rate, as a decimal.

The value of a certain car decreases by 16% each year.

This means that $r = 0.16$

$V(t) = V(0)(1-r)^{t}$

$V(t) = V(0)(1-0.16)^{t}$

$V(t) = V(0)(0.84)^{t}$

What is the 1⁄2-life of the car?

This is t for which V(t) = 0.5V(0). So

$V(t) = V(0)(0.84)^{t}$

$0.5V(0) = V(0)(0.84)^{t}$

$(0.84)^{t} = 0.5$

$\log{(0.84)^{t}} = \log{0.5}$

$t\log{0.84} = \log{0.5}$

$t = \frac{\log{0.5}}{\log{0.84}}$

$t = 3.98$

The half life of the car is 3.98 years.

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