y=a(1+r)^t
The variable y is the grizzly bear population, and t is time in years. We set that t=0 corresponds to five years ago. The variable r represents the annual rate of growth. Since we know that for t=0 the population was 325, let's substitute t=0 and y=325 into the equation.
Now we can partially write our equation.
y= a(1+r)^t ⇒ y= 325(1+r)^t
Since today the population is 450, we can substitute 450 for y and 5 for t. Let's do it!
We are asked to find how many years it will take to reach the maximum population, which is equal to 750. Therefore, we will substitute 750 for y into the equation from Part A and solve it for t.
To solve the equation for t, we will use the definition of a logarithm.
x= b^y ⇔ y=log_b x
This definition tells us how to rewrite the logarithm equivalent of y as an exponential equation. The argument x is equal to b raised to the power of y. Let's do it!
30/13= 1.067^t ⇔ t=log_(1.067) 30/13
Next, we will solve it for t.
