Question

A lake currently has a depth of 30 meters. As sediment builds up in the lake, its depth decreases by 2% per year. This situation represents . The rate of growth or decay, r, is equal to . So the depth of the lake each year is times the depth in the previous year. It will take between years for the depth of the lake to reach 26.7 meters.

Answer 1

Answer:

This situations represents the depth of the lake after t years.

The rate of decay is equal to 0.02 a year.

So the depth of the lake each year is 0.98 times the depth in the previous year.

It will take 5.77 years for the depth of the lake to reach 26.7 meters.

Step-by-step explanation:

Exponential equation of decay:

The exponential equation for the decay of an amount is given by:

$D(t) = D(0)(1-r)^t$

In which D(0) is the initial amount and r is the decay rate, as a decimal.

A lake currently has a depth of 30 meters. As sediment builds up in the lake, its depth decreases by 2% per year.

This means that $D(0) = 30, r = 0.02$

So

$D(t) = D(0)(1-r)^t$

$D(t) = 30(1-0.02)^t$

$D(t) = 30(0.98)^t$

This situation represents

The depth of the lake after t years.

The rate of growth or decay, r, is equal to

The rate of decay is equal to 0.02 a year.

So the depth of the lake each year is times the depth in the previous year.

1 - 0.02 = 0.9

So the depth of the lake each year is 0.98 times the depth in the previous year.

It will take between years for the depth of the lake to reach 26.7 meters.

This is t for which D(t) = 26.7. So

$D(t) = 30(0.98)^t$

$26.7 = 30(0.98)^t$

$(0.98)^t = \frac{26.7}{30}$

$\log{(0.98)^t} = \log{\frac{26.7}{30}}$

$t\log{(0.98)} = \log{\frac{26.7}{30}}$

$t = \frac{\log{\frac{26.7}{30}}}{\log{0.98}}$

$t = 5.77$

It will take 5.77 years for the depth of the lake to reach 26.7 meters.