A population is currently 400 and growing at a rate of 6% per year.

(a) Write a formula for the population P as a function of time t in years: P(t)= 400(1+.06)^t
(b) If the population continues this trend, what will it be in ten years? (Round off to the nearest whole person.)
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(c) If the population continues this trend, how many full years does it take to at least double?

a agree

b agree

2 = 1.06^n

n log 1.06 = log 2
n = log 2/ log 1.06
n = 11.9 so 12

To find out how many full years it takes for the population to at least double, we need to find when the population P(t) equals or exceeds 800. We can solve this using logarithms.

The formula for the population P(t) as a function of time t, which we derived earlier, is:

P(t) = 400(1 + 0.06)^t

Now, we can set up the equation:

400(1 + 0.06)^t ≥ 800

Divide both sides of the equation by 400:

(1 + 0.06)^t ≥ 2

Taking the natural logarithm of both sides:

ln((1 + 0.06)^t) ≥ ln(2)

Using the logarithmic property that ln(a^b) = b ln(a):

t ln(1 + 0.06) ≥ ln(2)

Now, we can isolate t by dividing both sides by ln(1 + 0.06):

t ≥ ln(2) / ln(1 + 0.06)

Using a calculator, we can find:

t ≥ 11.509
Round up to the nearest whole number to get:

t ≥ 12

Therefore, it will take at least 12 full years for the population to double if it continues growing at a rate of 6% per year.

To find out how many full years it takes for the population to at least double, we need to solve the inequality:

400(1+.06)^t ≥ 800

First, let's divide both sides of the inequality by 400:

(1+.06)^t ≥ 2

Next, take the natural log (ln) of both sides of the inequality:

ln((1+.06)^t) ≥ ln(2)

Using the property of logarithms that ln(a^b) = b ln(a), we can simplify it further:

t ln(1+.06) ≥ ln(2)

Divide both sides of the inequality by ln(1+.06):

t ≥ ln(2) / ln(1+.06)

Using a calculator, we can find that ln(2) is approximately 0.6931 and ln(1+.06) is approximately 0.0590. So the inequality becomes:

t ≥ 0.6931 / 0.0590

t ≥ 11.744

Therefore, it will take at least 12 years for the population to double.